The polynomial $p(x)$ is defined by $p(x) = 9x^3 + 18x^2 + 5x + 4$. (a) Find the quotient when $p(x)$ is divided by $(3x + 2)$, and show that the remainder is 6. (b) Find the value of $\int_0^2 \frac{p(x)}{3x + 2} \,dx$, giving your answer in the form $a + \ln b$ where $a$ and $b$ are integers.

(a) Quotient: 3x^2 + 4x - 1; remainder: 6 · (b) 14 + ln16

The diagram shows the curve with equation $y = \frac{\ln(2x + 1)}{x + 3}$. The curve has a maximum point $M$. (a) Find an expression for $\frac{dy}{dx}$. (b) Show that the $x$-coordinate of $M$ satisfies the equation $x = \frac{x + 3}{\ln(2x + 1)} - 0.5$. (c) Show by calculation that the $x$-coordinate of $M$ lies between 2.5 and 3.0. (d) Use an iterative formula based on the equation in part (b) to find the $x$-coordinate of $M$ correct to 4 significant figures. Give the…

(a) dy/dx = [(x+3)·(2/(2x+1)) - ln(2x+1)] / (x+3)² · (b) x = (x+3)/ln(2x+1) - 0.5 · (c) Sign change from -0.0696 to 0.4166 confirms the root is in [2.5, 3.0]. · (d) x = 2.569

9709/23 May/June 2024 — Mathematics Solved Past Paper

Q14MMediumAlgebra

Solve the inequality $|5x + 7| > |2x - 3|$ .

DifficultyMedium

Worked solution

Approach

To solve $|5x + 7| > |2x - 3|$ , we first find the critical values of $x$ where the two expressions are equal in magnitude. These values divide the number line into intervals. We then test a value from each interval to determine where the inequality holds.

Working

Solve the equality $|5x + 7| = |2x - 3|$ . This gives two linear equations:

5x + 7 = 2x - 3 \quad \text{or} \quad 5x + 7 = -(2x - 3)

Solving the first:

5x + 7 = 2x - 3 \implies 3x = -10 \implies x = -\frac{10}{3}

Solving the second:

5x + 7 = -2x + 3 \implies 7x = -4 \implies x = -\frac{4}{7}

The critical points are $x = -\frac{10}{3}$ and $x = -\frac{4}{7}$ . These divide the real line into three intervals: $x < -\frac{10}{3}$ , $-\frac{10}{3} < x < -\frac{4}{7}$ , and $x > -\frac{4}{7}$ .

Test a point in each interval:

For $x < -\frac{10}{3}$ , choose $x = -4$ : $|5(-4)+7| = 13$ and $|2(-4)-3| = 11$ . Since $13 > 11$ , the inequality holds.
For $-\frac{10}{3} < x < -\frac{4}{7}$ , choose $x = -1$ : $|5(-1)+7| = 2$ and $|2(-1)-3| = 5$ . Since $2 > 5$ is false, the inequality does not hold.
For $x > -\frac{4}{7}$ , choose $x = 0$ : $|5(0)+7| = 7$ and $|2(0)-3| = 3$ . Since $7 > 3$ , the inequality holds.

Therefore, the solution is the union of the two outer intervals.

Answer

x < -\frac{10}{3} \quad \text{or} \quad x > -\frac{4}{7}

Final answer

x < -10/3 or x > -4/7

Detailed explanation

Walkthrough

The modulus of a number is its distance from zero, so $|5x+7| > |2x-3|$ means the distance of $5x+7$ from zero is greater than the distance of $2x-3$ from zero. The boundary between "greater" and "less" occurs where the two distances are equal, i.e. $|5x+7| = |2x-3|$ . Solving this equality gives the critical points. Since the expressions are linear, the inequality changes behaviour only at these points, so we test one value in each interval.

Alternatively, since both sides are non-negative, we can square both sides: $(5x+7)^2 > (2x-3)^2$ , which simplifies to a quadratic inequality. Expanding gives $25x^2 + 70x + 49 > 4x^2 - 12x + 9$ , so $21x^2 + 82x + 40 > 0$ . Factorising gives $(3x+10)(7x+4) > 0$ . The roots are $-\frac{10}{3}$ and $-\frac{4}{7}$ . Because the quadratic opens upwards, the inequality holds outside the roots, giving the same solution $x < -\frac{10}{3}$ or $x > -\frac{4}{7}$ .

Key Takeaways

The critical point method for modulus inequalities: solve the equality to find boundaries, then test intervals.
The squaring method: squaring both sides of a modulus inequality is valid because both sides are non-negative, converting it into a standard quadratic inequality.
Understanding that the solution to such inequalities is often a union of two disjoint intervals, expressed with "or".

Common Mistakes

Using "and" instead of "or" when writing the final answer. The mark scheme explicitly states A0 if '... and ...' is used.
Forgetting to test the intervals after finding the critical points, leading to an incorrect final answer.
Arithmetic errors when solving the linear equations, especially with the negative sign in $5x+7 = -(2x-3)$ .
When squaring, incorrectly expanding $(2x-3)^2$ (forgetting the $-12x$ term).

Things to Be Careful About

The inequality is strict ( $>$ ), so the critical points themselves are not included in the solution set. The final answer must use strict inequalities.
When testing intervals, choose convenient integer values that clearly fall into each interval.
Ensure the final answer is written as two separate inequalities joined by "or", not "and".

Techniques used

find critical values by solving the modulus equalitytest intervals to determine where the inequality holdsstate the solution as two disjoint inequalities

Approach

Take natural logarithms of both sides so the unknown $x$ is brought down from the exponents. Use the power rule for logarithms on the left and the fact that $\ln e = 1$ on the right, then solve the resulting linear equation.

Working

Take $\ln$ of both sides:

(2x - 1)\ln 6 = \ln\left(5e^{3x+2}\right)

Using $\ln(ab) = \ln a + \ln b$ and $\ln e = 1$ :

(2x - 1)\ln 6 = \ln 5 + 3x + 2

Expand the left-hand side:

2x\ln 6 - \ln 6 = \ln 5 + 3x + 2

Collect the $x$ -terms on one side:

2x\ln 6 - 3x = \ln 5 + 2 + \ln 6

Factor out $x$ :

x(2\ln 6 - 3) = \ln 5 + \ln 6 + 2

Hence:

x = \frac{\ln 5 + \ln 6 + 2}{2\ln 6 - 3}

Evaluating:

x = \frac{1.609437912 + 1.791759469 + 2}{2(1.791759469) - 3} = \frac{5.401197381}{0.583518938} = 9.256

Answer

x = 9.256

Final answer

x = 9.256

Detailed explanation

Walkthrough

We need to solve an equation in which $x$ appears in the exponents. Since logarithms are the inverse operation of exponentials, taking natural logarithms of both sides is the natural first step. The bases are different ( $6$ and $e$ ), so we cannot simply equate the exponents; taking logs allows each exponent to be brought down.

Taking $\ln$ of both sides gives:

(2x - 1)\ln 6 = \ln\left(5e^{3x+2}\right)

On the left, the power rule for logarithms lets us move the exponent $2x - 1$ in front of $\ln 6$ . On the right, the product rule gives $\ln 5 + \ln\left(e^{3x+2}\right)$ , and since $\ln e = 1$ , this becomes $\ln 5 + 3x + 2$ .

Expanding and rearranging turns the equation into a linear equation in $x$ :

2x\ln 6 - 3x = \ln 5 + \ln 6 + 2

Factorising $x$ and dividing by its coefficient gives the exact expression for $x$ . Finally, substituting the values of $\ln 5$ and $\ln 6$ and evaluating gives $x = 9.256$ , correct to 4 significant figures.

The mark scheme awards the first two method marks for writing the two logarithmic forms, a dependent method mark for attempting to solve the linear equation, and the final accuracy mark for $9.256$ .

Key Takeaways

Logarithms convert equations with the unknown in an exponent into linear equations.
The power rule $\ln(a^b) = b\ln a$ is essential for bringing exponents down.
The product rule $\ln(ab) = \ln a + \ln b$ and the identity $\ln e = 1$ are needed to simplify the right-hand side.
Always give the final answer to the required number of significant figures.

Common Mistakes

Forgetting to take the logarithm of the whole right-hand side, so the product $5e^{3x+2}$ is not expanded correctly.
Writing $\ln\left(5e^{3x+2}\right)$ as $\ln 5 \cdot \ln\left(e^{3x+2}\right)$ instead of $\ln 5 + \ln\left(e^{3x+2}\right)$ .
Forgetting that $\ln e = 1$ , leaving an unnecessary $\ln e$ term.
Making sign errors when moving terms such as $-\ln 6$ or $3x$ across the equation.
Rounding intermediate values too early, which can change the final digit.

Things to Be Careful About

The right-hand side is a product, so use $\ln(ab) = \ln a + \ln b$ , not $\ln a \cdot \ln b$ .
Since $\ln e = 1$ , the term $3x + 2$ appears directly without a logarithm.
Show the linear equation and its rearrangement to earn the dependent method mark.
The final answer must be given to 4 significant figures: $9.256$ . Greater accuracy such as $9.2562$ or $9.25625$ is also accepted by the mark scheme.

Techniques used

take natural logarithms of both sidesapply the power rule for logarithmsuse the identity ln e = 1collect x terms and solve the resulting linear equation

Approach

To find the x-coordinate of B, set $y = 0$ and solve for $x$ . To find the area of the shaded region, integrate the curve equation from $x = 0$ to the x-coordinate of B.

Working

Finding the x-coordinate of B:

At point B, $y = 0$ :

8e^{-x} - e^{2x} = 0

8e^{-x} = e^{2x}

Multiply both sides by $e^x$ :

8 = e^{3x}

Take the natural logarithm of both sides:

3x = \ln 8

Since $8 = 2^3$ , we have $\ln 8 = 3\ln 2$ :

3x = 3\ln 2

x = \ln 2

This shows that the x-coordinate of B is $\ln 2$ .

Finding the area of the shaded region:

The area is given by the definite integral:

\text{Area} = \int_0^{\ln 2} \left(8e^{-x} - e^{2x}\right) \, dx

Integrate:

\left[ -8e^{-x} - \frac{1}{2}e^{2x} \right]_0^{\ln 2}

Evaluate at the upper limit $x = \ln 2$ :

-8e^{-\ln 2} - \frac{1}{2}e^{2\ln 2} = -8\left(\frac{1}{2}\right) - \frac{1}{2}\left(2^2\right) = -4 - 2 = -6

Evaluate at the lower limit $x = 0$ :

-8e^{0} - \frac{1}{2}e^{0} = -8 - \frac{1}{2} = -\frac{17}{2}

Subtract the lower limit value from the upper limit value:

\text{Area} = -6 - \left(-\frac{17}{2}\right) = -6 + \frac{17}{2} = \frac{5}{2}

Answer

\frac{5}{2}

Final answer

5/2

Detailed explanation

Walkthrough

First, find where the curve crosses the x-axis by setting $y = 0$ . This gives $8e^{-x} = e^{2x}$ . Multiplying by $e^x$ simplifies this to $e^{3x} = 8$ . Taking the natural log gives $3x = \ln 8 = 3\ln 2$ , so $x = \ln 2$ . Next, integrate the function from $x = 0$ to $x = \ln 2$ . The antiderivative is $-8e^{-x} - \frac{1}{2}e^{2x}$ . Evaluating at $\ln 2$ requires knowing that $e^{-\ln 2} = \frac{1}{2}$ and $e^{2\ln 2} = 4$ . Evaluating at 0 gives $-8 - 0.5 = -8.5$ . The difference is $-6 - (-8.5) = 2.5$ .

Key Takeaways

Exponential equations can be solved by isolating the exponential term and using logarithms.
Definite integrals with exponential functions require careful evaluation of limits, especially when limits involve logarithms.

Common Mistakes

Failing to simplify $e^{2\ln 2}$ correctly as $e^{\ln 4} = 4$ .
Sign errors when subtracting the lower limit evaluation from the upper limit evaluation.
Forgetting the chain rule factor when integrating $e^{2x}$ (must divide by 2).

Things to Be Careful About

Ensure the limits of integration are applied correctly: upper limit minus lower limit.
Remember that $e^{-\ln a} = \frac{1}{a}$ and $e^{k\ln a} = a^k$ .

Techniques used

solve exponential equationintegrate exponential functionsevaluate definite integral with logarithmic limits

Q47MMediumDifferentiation

A curve is defined by the parametric equations

x = 4\cos^2 t, \quad y = \sqrt{3}\sin 2t,

for values of $t$ such that $0 < t < \frac{1}{2}\pi$ .

Find the equation of the normal to the curve at the point for which $t = \frac{1}{6}\pi$ . Give your answer in the form $ax + by + c = 0$ where $a$ , $b$ and $c$ are integers.

DifficultyMedium

Worked solution

Approach

Differentiate $x$ and $y$ with respect to the parameter $t$ , then use

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

Evaluate at $t = \frac{\pi}{6}$ , take the negative reciprocal for the normal gradient, and form the equation of the normal through the corresponding point.

Working

Differentiate $x = 4\cos^2 t$ :

\frac{dx}{dt} = -8\cos t \sin t = -4\sin 2t

Differentiate $y = \sqrt{3}\sin 2t$ :

\frac{dy}{dt} = 2\sqrt{3}\cos 2t

At $t = \frac{\pi}{6}$ :

\frac{dx}{dt} = -8\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) = -2\sqrt{3}, \qquad \frac{dy}{dt} = 2\sqrt{3}\cos\frac{\pi}{3} = \sqrt{3}

Therefore

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\sqrt{3}}{-2\sqrt{3}} = -\frac{1}{2}

The gradient of the normal is the negative reciprocal:

m_{\text{normal}} = 2

The point on the curve when $t = \frac{\pi}{6}$ is

x = 4\cos^2\frac{\pi}{6} = 3, \qquad y = \sqrt{3}\sin\frac{\pi}{3} = \frac{3}{2}

So the normal has equation

y - \frac{3}{2} = 2(x - 3)

Multiplying by 2 and rearranging:

2y - 3 = 4x - 12 \quad \Rightarrow \quad 4x - 2y - 9 = 0

Answer

4x - 2y - 9 = 0

Final answer

4x - 2y - 9 = 0

Detailed explanation

Walkthrough

The curve is given in parametric form, so we cannot write $y$ directly as a function of $x$ . Instead, we differentiate both $x$ and $y$ with respect to the parameter $t$ , then use the chain rule relation

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.

This is the key idea of parametric differentiation.

First differentiate $x = 4\cos^2 t$ . Using the chain rule, the derivative of $\cos^2 t$ is $2\cos t \cdot (-\sin t)$ , so

\frac{dx}{dt} = -8\cos t \sin t = -4\sin 2t.

The second form uses the double-angle identity $\sin 2t = 2\sin t \cos t$ .

Next differentiate $y = \sqrt{3}\sin 2t$ . Again by the chain rule,

\frac{dy}{dt} = 2\sqrt{3}\cos 2t.

Now substitute $t = \frac{\pi}{6}$ into both derivatives. We have

\frac{dx}{dt} = -8\cdot\frac{\sqrt{3}}{2}\cdot\frac{1}{2} = -2\sqrt{3}

and

\frac{dy}{dt} = 2\sqrt{3}\cos\frac{\pi}{3} = \sqrt{3}.

Therefore the gradient of the tangent is

\frac{dy}{dx} = \frac{\sqrt{3}}{-2\sqrt{3}} = -\frac{1}{2}.

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal:

m_{\text{normal}} = 2.

We also need the coordinates of the point on the curve. Substitute $t = \frac{\pi}{6}$ into the original parametric equations:

x = 4\cos^2\frac{\pi}{6} = 4\cdot\frac{3}{4} = 3, \qquad y = \sqrt{3}\sin\frac{\pi}{3} = \sqrt{3}\cdot\frac{\sqrt{3}}{2} = \frac{3}{2}.

Using the point-gradient form of a straight line,

y - \frac{3}{2} = 2(x - 3).

Multiplying through by 2 and rearranging gives

4x - 2y - 9 = 0.

Key Takeaways

Parametric differentiation uses $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ .
The chain rule is essential for differentiating composite trigonometric functions such as $\cos^2 t$ and $\sin 2t$ .
The gradient of the normal is the negative reciprocal of the gradient of the tangent.
Always find the point of contact by substituting the parameter value into the original equations.
Present the final equation in the required form $ax + by + c = 0$ .

Common Mistakes

Forgetting the negative sign when differentiating $\cos t$ : $\frac{d}{dt}\cos t = -\sin t$ .
Writing $\frac{dy}{dx}$ as $\frac{dx/dt}{dy/dt}$ instead of the reciprocal.
Using the tangent gradient instead of the normal gradient.
Substituting $t$ into the derivative formula but not into the original equations to find the point.
Leaving the answer as $y - \frac{3}{2} = 2(x - 3)$ instead of converting to the requested $ax + by + c = 0$ form.

Things to Be Careful About

The domain $0 < t < \frac{\pi}{2}$ means $t = \frac{\pi}{6}$ is valid, and all trig values used are standard.
At $t = \frac{\pi}{6}$ , remember $\cos\frac{\pi}{3} = \frac{1}{2}$ and $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ .
The mark scheme requires a visible attempt to substitute $t$ into the derivatives before stating $\frac{dy}{dx} = -\frac{1}{2}$ .
The final equation may be multiplied by a constant and still be correct, but the requested form asks for integer coefficients.
Be careful with the sign of $\frac{dx}{dt}$ ; a sign error here changes the gradient and the final normal equation.

Techniques used

differentiate parametric equations with respect to the parameterapply the chain rule to composite trigonometric functionsevaluate the derivative at a specific parameter valueuse the negative reciprocal to find the normal gradientform the equation of a straight line through a given point

Approach

Divide $9x^3+18x^2+5x+4$ by $3x+2$ using polynomial long division, matching leading terms until the remainder has degree less than the divisor.

Working

First term:

\frac{9x^3}{3x}=3x^2

Multiply and subtract:

9x^3+18x^2+5x+4-3x^2(3x+2)=12x^2+5x+4

Next term:

\frac{12x^2}{3x}=4x

Subtract:

12x^2+5x+4-4x(3x+2)=-3x+4

Next term:

\frac{-3x}{3x}=-1

Subtract:

-3x+4-(-1)(3x+2)=6

Thus

p(x)=(3x+2)(3x^2+4x-1)+6

So the quotient is $3x^2+4x-1$ and the remainder is $6$ .

Answer

Quotient: $3x^2+4x-1$ ; remainder: $6$ .

Final answer

Quotient: 3x^2 + 4x - 1; remainder: 6

Detailed explanation

Walkthrough

We need to divide $p(x)$ by $3x+2$ . Polynomial long division works like ordinary division: divide the leading term of the dividend by the leading term of the divisor, multiply back, and subtract. Repeating this removes the $x^3$ , then the $x^2$ , then the $x$ terms, leaving a constant remainder.

First, $9x^3 \div 3x = 3x^2$ . Multiplying $3x^2(3x+2)$ gives $9x^3+6x^2$ , and subtracting leaves $12x^2+5x+4$ . Next, $12x^2 \div 3x = 4x$ . Multiplying and subtracting leaves $-3x+4$ . Finally, $-3x \div 3x = -1$ . Multiplying and subtracting leaves $6$ . Since $6$ has degree less than the divisor $3x+2$ , the division stops. The quotient is $3x^2+4x-1$ and the remainder is $6$ .

Key Takeaways

This question tests polynomial division by a linear factor and the relationship $p(x) = (3x+2)q(x) + r$ . It also shows that the remainder is a constant because the divisor is linear.

Common Mistakes

Stopping too early and not carrying the division down to the constant term.
Sign errors when subtracting the product of the divisor and the current quotient term.
Forgetting that the question says the remainder is given as 6, so the working must include enough detail to justify that remainder.

Things to Be Careful About

The divisor is $3x+2$ , not $x+2$ , so the leading terms are divided by $3x$ , not by $x$ . The remainder is the final constant left after the degree of the remainder is less than the degree of the divisor.

Techniques used

divide a polynomial by a linear divisormatch leading termssubtract products iterativelyidentify the quotient and remainder

(b)

Find the value of $\int_0^2 \frac{p(x)}{3x + 2} \,dx$ , giving your answer in the form $a + \ln b$ where $a$ and $b$ are integers.

5M Similar

DifficultyMedium

Worked solution

Approach

Use the quotient and remainder from part (a) to rewrite the integrand as a polynomial plus a fractional term. Then integrate each term separately, using $\int \frac{1}{3x+2}\,dx = \frac{1}{3}\ln(3x+2)$ , and evaluate between the limits.

Working

From part (a):

\frac{p(x)}{3x+2}=3x^2+4x-1+\frac{6}{3x+2}

Therefore

\int_0^2 \frac{p(x)}{3x+2}\,dx=\int_0^2\left(3x^2+4x-1+\frac{6}{3x+2}\right)dx

Integrate term by term:

\left[x^3+2x^2-x+2\ln(3x+2)\right]_0^2

At $x=2$ :

2^3+2(2)^2-2+2\ln(8)=8+8-2+2\ln8=14+2\ln8

At $x=0$ :

0^3+2(0)^2-0+2\ln(2)=2\ln2

Subtract the lower limit from the upper limit:

14+2\ln8-2\ln2=14+2\ln\left(\frac{8}{2}\right)=14+2\ln4=14+\ln16

Answer

14+\ln16

Final answer

14 + ln16

Detailed explanation

Walkthrough

Part (a) gives $p(x)=(3x+2)(3x^2+4x-1)+6$ . Dividing by $3x+2$ therefore gives $3x^2+4x-1+\frac{6}{3x+2}$ . This is the key step: it turns a difficult-looking cubic-over-linear integrand into terms we can integrate directly.

Integrate each term: $\int 3x^2\,dx=x^3$ , $\int 4x\,dx=2x^2$ , $\int -1\,dx=-x$ , and $\int \frac{6}{3x+2}\,dx=2\ln(3x+2)$ because the derivative of $\ln(3x+2)$ is $\frac{3}{3x+2}$ .

Then evaluate at $x=2$ and $x=0$ . The upper limit gives $14+2\ln8$ ; the lower limit gives $2\ln2$ . Subtracting gives $14+2\ln8-2\ln2$ . Use the logarithm law $\ln A-\ln B=\ln\frac{A}{B}$ to combine $2\ln8-2\ln2=2\ln4=\ln16$ . Hence the value is $14+\ln16$ .

Key Takeaways

This question combines polynomial division with integration. It also reinforces that $\int \frac{k}{ax+b}\,dx=\frac{k}{a}\ln|ax+b|$ and that logarithm laws are needed when evaluating definite integrals with logarithmic terms.

Common Mistakes

Forgetting to divide the fractional coefficient correctly: $\int \frac{6}{3x+2}\,dx=2\ln(3x+2)$ , not $6\ln(3x+2)$ .
Forgetting to subtract the lower-limit value when evaluating the definite integral.
Incorrectly combining logarithms, e.g. writing $2\ln8-2\ln2$ as $\ln(8-2)$ instead of using the quotient law.
Not using the result from part (a) and trying to integrate the original fraction directly.

Things to Be Careful About

The integral of $\frac{1}{3x+2}$ is $\frac{1}{3}\ln(3x+2)$ , so the factor $6$ gives $2\ln(3x+2)$ . When substituting $x=0$ , the logarithmic term is $2\ln2$ , not $0$ . Finally, express the answer in the required form $a+\ln b$ with integers $a$ and $b$ ; here $a=14$ and $b=16$ .

Techniques used

rewrite the integrand using the polynomial division resultintegrate polynomial and reciprocal termsapply logarithm laws to combine termsevaluate a definite integral with limits

Q6MediumDifferentiationNumerical Solution of Equations

Fig. 2

The diagram shows the curve with equation $y = \frac{\ln(2x + 1)}{x + 3}$ . The curve has a maximum point $M$ .

(a)

Find an expression for $\frac{dy}{dx}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

We differentiate $y = \frac{\ln(2x+1)}{x+3}$ using the quotient rule, with the chain rule supplying the derivative of $\ln(2x+1)$ .

Working

Identify $u = \ln(2x+1)$ and $v = x+3$ . By the quotient rule,

\frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2}

By the chain rule, $u' = \frac{2}{2x+1}$ , and $v' = 1$ . Substituting,

\frac{dy}{dx} = \frac{(x+3) \cdot \frac{2}{2x+1} - \ln(2x+1) \cdot 1}{(x+3)^2}

Answer

\frac{dy}{dx} = \frac{(x+3) \cdot \frac{2}{2x+1} - \ln(2x+1)}{(x+3)^2}

Final answer

dy/dx = [(x+3)·(2/(2x+1)) - ln(2x+1)] / (x+3)²

Detailed explanation

Walkthrough

The function $y = \frac{\ln(2x+1)}{x+3}$ is a quotient of two functions of $x$ , so the quotient rule is the natural tool. Letting $u = \ln(2x+1)$ (the numerator) and $v = x+3$ (the denominator), the quotient rule states $\frac{dy}{dx} = \frac{v u' - u v'}{v^2}$ .

The derivative $v' = 1$ is immediate, but $u'$ requires the chain rule. Since $\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$ and $g(x) = 2x+1$ has $g'(x) = 2$ , we have $u' = \frac{2}{2x+1}$ . The factor of 2 is the chain-rule contribution; omitting it would give the wrong answer.

Substituting $u$ , $u'$ , $v$ , $v'$ into the quotient-rule formula gives the required expression. This derivative is the foundation for parts (b), (c) and (d).

Key Takeaways

The quotient rule is the standard tool for differentiating a fraction of two functions of $x$ .
The chain rule must be applied when differentiating $\ln(g(x))$ , giving $\frac{g'(x)}{g(x)}$ , not just $\frac{1}{g(x)}$ .
The expression for $\frac{dy}{dx}$ is needed in part (b) to locate the maximum.

Common Mistakes

Forgetting the chain-rule factor of 2 when differentiating $\ln(2x+1)$ , giving $\frac{1}{2x+1}$ instead of $\frac{2}{2x+1}$ .
Reversing the order in the quotient rule (it is $v u' - u v'$ , not $u v' - v u'$ ).
Forgetting to square the denominator.

Things to Be Careful About

The denominator $(x+3)^2$ is non-negative and positive for $x \neq -3$ , so the sign of $\frac{dy}{dx}$ is determined by the numerator alone — useful in part (b).

Techniques used

apply quotient rule for differentiationapply chain rule to ln(2x+1)

(b)

Show that the $x$ -coordinate of $M$ satisfies the equation $x = \frac{x + 3}{\ln(2x + 1)} - 0.5$ .

2M Similar

DifficultyMedium

Worked solution

Approach

At the maximum point $M$ , $\frac{dy}{dx} = 0$ . We set the numerator of the expression from part (a) equal to zero and rearrange into the required form.

Working

Setting $\frac{dy}{dx} = 0$ and clearing the positive denominator $(x+3)^2$ :

(x+3) \cdot \frac{2}{2x+1} - \ln(2x+1) = 0

Rearrange:

(x+3) \cdot \frac{2}{2x+1} = \ln(2x+1)

Multiply both sides by $(2x+1)$ :

2(x+3) = (2x+1)\,\ln(2x+1)

Divide both sides by $\ln(2x+1)$ (which is positive for $x > 0$ ):

\frac{2(x+3)}{\ln(2x+1)} = 2x+1

This is the $2x+1 = \ldots$ form identified in the mark scheme. Divide both sides by $2$ :

\frac{x+3}{\ln(2x+1)} = \frac{2x+1}{2} = x + 0.5

Rearrange to isolate $x$ :

x = \frac{x+3}{\ln(2x+1)} - 0.5

Answer

x = \frac{x+3}{\ln(2x+1)} - 0.5

Final answer

x = (x+3)/ln(2x+1) - 0.5

Detailed explanation

Walkthrough

The maximum point $M$ is a turning point, so its gradient is zero. We use the result from part (a) and set the numerator of $\frac{dy}{dx}$ to zero. (The denominator $(x+3)^2$ is positive in the relevant range, so the sign of $\frac{dy}{dx}$ depends only on the numerator.)

Starting from
$(x+3) \cdot \frac{2}{2x+1} = \ln(2x+1)$
we multiply by $(2x+1)$ to give $2(x+3) = (2x+1)\ln(2x+1)$ , then divide by $\ln(2x+1)$ to isolate $2x+1$ :
$\frac{2(x+3)}{\ln(2x+1)} = 2x+1.$
This is the $2x+1 = \ldots$ form the mark scheme identifies. To reach the required form, divide both sides by $2$ to obtain $\frac{x+3}{\ln(2x+1)} = x + 0.5$ , then rearrange to
$x = \frac{x+3}{\ln(2x+1)} - 0.5.$

Key Takeaways

A maximum point occurs where the derivative is zero.
Algebraic manipulation can convert the critical-point equation into an iteration-friendly form $x = f(x)$ .
The form derived here becomes the iteration formula in part (d).

Common Mistakes

Forgetting to clear the denominator or to divide by the leading coefficient $2$ .
Dividing by $\ln(2x+1)$ when it is non-positive (for $x > 0$ the logarithm is positive, so this is safe).
Sign errors when isolating $x$ on the left-hand side.

Things to Be Careful About

The denominator $\ln(2x+1)$ must be positive for the rearrangement to be valid; this holds for $x > 0$ , which includes the region of interest.
The required form is the iteration formula; keep $0.5$ (not $\frac{1}{2}$ ) on the right.

Techniques used

set derivative equal to zero for maximumrearrange equation algebraicallydivide by coefficient to isolate x

(c)

Show by calculation that the $x$ -coordinate of $M$ lies between 2.5 and 3.0.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Define $g(x) = x - \frac{x+3}{\ln(2x+1)} + 0.5$ , which is zero precisely at the $x$ -coordinate of $M$ . We evaluate $g$ at the endpoints of $[2.5, 3.0]$ and look for a sign change.

Working

At $x = 2.5$ :

g(2.5) = 2.5 - \frac{5.5}{\ln 6} + 0.5 = 3 - \frac{5.5}{1.79176} = 3 - 3.0696 = -0.0696

At $x = 3$ :

g(3) = 3 - \frac{6}{\ln 7} + 0.5 = 3.5 - \frac{6}{1.94591} = 3.5 - 3.0834 = 0.4166

Since $g(2.5) = -0.0696 < 0$ and $g(3) = 0.4166 > 0$ , the continuous function $g$ changes sign on $[2.5, 3.0]$ . By the intermediate value theorem, $g$ has a root in this interval, so the $x$ -coordinate of $M$ lies between 2.5 and 3.0.

Answer

The $x$ -coordinate of $M$ lies between 2.5 and 3.0 (sign change from $-0.0696$ to $0.4166$ is confirmed).

Final answer

Sign change from -0.0696 to 0.4166 confirms the root is in [2.5, 3.0].

Detailed explanation

Walkthrough

To locate the root of the equation from part (b), we use the sign-change method. Define
$g(x) = x - \frac{x+3}{\ln(2x+1)} + 0.5.$
This function is zero exactly when $x = \frac{x+3}{\ln(2x+1)} - 0.5$ , i.e., at the $x$ -coordinate of $M$ .

Evaluate $g$ at the endpoints of the candidate interval $[2.5, 3.0]$ :

$g(2.5) = 2.5 - \frac{5.5}{\ln 6} + 0.5 = 3 - 3.0696 = -0.0696$ . Negative.
$g(3) = 3 - \frac{6}{\ln 7} + 0.5 = 3.5 - 3.0834 = 0.4166$ . Positive.

The function $g$ is continuous on $[2.5, 3.0]$ (since $\ln(2x+1) > 0$ for $x > 0$ ). By the intermediate value theorem, because $g$ changes sign across the interval, it must have a root in $[2.5, 3.0]$ . Hence the $x$ -coordinate of $M$ lies in this interval.

Alternative view. Let $f(x) = \frac{x+3}{\ln(2x+1)} - 0.5$ . Then $f(2.5) = 2.5696 < 3$ and $f(3) = 2.5834 > 2.5$ , so the iteration $x_{n+1} = f(x_n)$ maps the interval $[2.5, 3.0]$ back into itself. By the fixed-point principle, $f$ has a fixed point in the interval, which is the desired $x$ -coordinate of $M$ .

Key Takeaways

A sign change of a continuous function across an interval implies a root in the interval (intermediate value theorem).
Working with $g(x) = x - f(x)$ turns a fixed-point problem into a root-locating problem.
The interval identified here is the starting range for the iteration in part (d).

Common Mistakes

Using the wrong sign in $g(x)$ : it must be $x - f(x)$ , not $f(x) - x$ , so that $g(x) = 0 \iff f(x) = x$ .
Computational slips in evaluating $\ln 6 \approx 1.79176$ or $\ln 7 \approx 1.94591$ .
Stating numerical values without the explicit sign-change conclusion.

Things to Be Careful About

The function $g$ is continuous wherever $\ln(2x+1) \neq 0$ , i.e., $x \neq -0.5$ . On $[2.5, 3.0]$ this is fine.
The unrounded values ( $-0.0696$ and $0.4166$ ) are required to support the sign-change argument; the rounded values $-0.07$ and $0.4$ are insufficient because the rounding could flip the sign.

Techniques used

rearrange equation into a function equal to zerosubstitute endpoint valuesapply sign change to locate root

(d)

Use an iterative formula based on the equation in part (b) to find the $x$ -coordinate of $M$ correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the iteration $x_{n+1} = \frac{x_n+3}{\ln(2x_n+1)} - 0.5$ from part (b), starting from $x_0 = 2.5$ (a value in the interval established in part (c)). Iterate to 6 sf until consecutive values agree to the required precision.

Working

With $x_0 = 2.5$ :

Iteration 1

x_1 = \frac{2.5+3}{\ln(2(2.5)+1)} - 0.5 = \frac{5.5}{\ln 6} - 0.5 = \frac{5.5}{1.79176} - 0.5 = 3.06961 - 0.5 = 2.56961

Iteration 2

x_2 = \frac{2.56961+3}{\ln(2(2.56961)+1)} - 0.5 = \frac{5.56961}{\ln 6.13922} - 0.5 = \frac{5.56961}{1.81470} - 0.5 = 3.06917 - 0.5 = 2.56917

Iteration 3

x_3 = \frac{2.56917+3}{\ln(2(2.56917)+1)} - 0.5 = \frac{5.56917}{\ln 6.13834} - 0.5 = \frac{5.56917}{1.81455} - 0.5 = 3.06917 - 0.5 = 2.56917

The values $x_2 = 2.56917$ and $x_3 = 2.56917$ both lie in the interval $[2.5685, 2.5695]$ and agree to 6 significant figures, demonstrating convergence. The $x$ -coordinate of $M$ correct to 4 significant figures is $2.569$ .

Answer

x = 2.569

Final answer

x = 2.569

Detailed explanation

Walkthrough

The equation from part (b), $x = \frac{x+3}{\ln(2x+1)} - 0.5$ , defines a fixed-point iteration $x_{n+1} = F(x_n)$ where $F(x) = \frac{x+3}{\ln(2x+1)} - 0.5$ . Starting from a value in the interval identified in part (c), we generate successive approximations.

Begin with $x_0 = 2.5$ . Each iteration:

Compute $2x_n + 1$ , then $\ln(2x_n+1)$ .
Compute $x_n + 3$ , divide by the logarithm, and subtract $0.5$ .

Iteration 1: $x_1 = \frac{5.5}{\ln 6} - 0.5 = \frac{5.5}{1.79176} - 0.5 = 2.56961$ .

Iteration 2: $x_2 = \frac{5.56961}{\ln 6.13922} - 0.5 = \frac{5.56961}{1.81470} - 0.5 = 2.56917$ .

Iteration 3: $x_3 = \frac{5.56917}{\ln 6.13834} - 0.5 = \frac{5.56917}{1.81455} - 0.5 = 2.56917$ .

The values $x_2$ and $x_3$ both lie in $[2.5685, 2.5695]$ and agree to 6 significant figures, so the iteration has converged to the required accuracy. The 4 sf answer is $x = 2.569$ .

(Equivalently, $F'(2.5) \approx -0.013$ and $F'(3) \approx 0.061$ , both of magnitude less than 1, so the iteration is a contraction on $[2.5, 3.0]$ and converges quickly to the unique fixed point.)

Key Takeaways

The fixed-point iteration $x_{n+1} = F(x_n)$ converges to the root when started within the interval of convergence.
Sufficient iterations to higher precision are required to justify a lower-precision final answer.
A common stopping criterion is to check that consecutive values lie in a narrow interval bracketing the root, or that they agree to the required precision.

Common Mistakes

Using the wrong iteration formula (e.g., a rearrangement that doesn't match the equation in part (b)).
Insufficient iterations: only one or two iterations don't justify 4 sf accuracy.
Premature rounding: each iteration should be carried to 6 sf or more so the final 4 sf answer is correct.

Things to Be Careful About

The iteration converges quickly here, but at least three iterations are needed to verify the 4 sf answer.
The sign of $\ln(2x+1)$ stays positive throughout (true for $x > 0$ ), so no iteration fails.
Final answer must be to exactly 4 sf: $2.569$ , not $2.5692$ (5 sf) or $2.57$ (3 sf).

Techniques used

apply iterative formula from part (b)compute successive iterations to 6 sfidentify convergence and round to 4 sf

Q7MediumTrigonometryIntegration

(a)

Prove that $2\sin\theta \csc 2\theta \equiv \sec\theta$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Rewrite $\csc 2\theta$ using the definition of cosecant, then apply the double-angle identity for sine and simplify to show the result equals $\sec\theta$ .

Working

Start with the left-hand side. Since $\csc 2\theta = \frac{1}{\sin 2\theta}$ :

2\sin\theta \csc 2\theta = 2\sin\theta \cdot \frac{1}{\sin 2\theta} = \frac{2\sin\theta}{\sin 2\theta}

Apply the double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$ :

\frac{2\sin\theta}{\sin 2\theta} = \frac{2\sin\theta}{2\sin\theta\cos\theta} = \frac{1}{\cos\theta}

Since $\frac{1}{\cos\theta} = \sec\theta$ , the identity is proved:

2\sin\theta \csc 2\theta = \sec\theta

Answer

2\sin\theta \csc 2\theta \equiv \sec\theta

Final answer

2 sin θ csc 2θ = sec θ

Detailed explanation

Walkthrough

We need to prove the identity $2\sin\theta \csc 2\theta = \sec\theta$ . Since the answer is given, we must show every step of the working.

Step 1 — Rewrite the cosecant. Recall that $\csc 2\theta = \frac{1}{\sin 2\theta}$ . This converts the reciprocal function into a fraction involving only sine.

Step 2 — Substitute into the LHS. The left-hand side becomes $2\sin\theta \cdot \frac{1}{\sin 2\theta} = \frac{2\sin\theta}{\sin 2\theta}$ .

Step 3 — Apply the double-angle identity. The denominator contains $\sin 2\theta$ , which we expand as $2\sin\theta\cos\theta$ . This is the crucial step: it introduces the $\cos\theta$ we need in the final answer.

Step 4 — Cancel common factors. Both numerator and denominator contain the factor $2\sin\theta$ , which cancels, leaving $\frac{1}{\cos\theta}$ .

Step 5 — Recognise secant. By definition $\sec\theta = \frac{1}{\cos\theta}$ , so the expression equals $\sec\theta$ , matching the right-hand side. The identity is proved.

Key Takeaways

The definitions $\csc\theta = \frac{1}{\sin\theta}$ and $\sec\theta = \frac{1}{\cos\theta}$ .
The double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$ .
The general strategy for proving trig identities: convert everything to sines and cosines, then simplify.

Common Mistakes

Writing $\csc 2\theta$ incorrectly as $\frac{1}{\sin\theta}$ instead of $\frac{1}{\sin 2\theta}$ .
Misapplying the double-angle formula, e.g. writing $\sin 2\theta = \sin\theta\cos\theta$ without the factor 2.
Because the answer is given, the mark scheme requires full working — skipping steps loses the marks.

Things to Be Careful About

The cancellation of $2\sin\theta$ is valid wherever the expression is defined (i.e. where $\sin\theta \neq 0$ and $\sin 2\theta \neq 0$ ). As an identity, we work on the common domain.
Present the working as a chain of equalities from the LHS to the RHS so the proof is clear.

Techniques used

express cosecant as reciprocal of sineapply the double-angle identity for sinecancel common factorsrecognise the reciprocal of cosine as secant

(b)

Solve the equation $\tan^2\theta + 7\sin\theta \csc 2\theta = 8$ for $-\pi < \theta < \pi$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use the identity proved in part (a) to replace $\sin\theta \csc 2\theta$ with $\frac{1}{2}\sec\theta$ . Then apply the Pythagorean identity $\tan^2\theta = \sec^2\theta - 1$ to obtain a quadratic equation in $\sec\theta$ . Solve for $\sec\theta$ , convert to $\cos\theta$ , and find all solutions in $-\pi < \theta < \pi$ .

Working

From part (a), $2\sin\theta \csc 2\theta = \sec\theta$ , so:

\sin\theta \csc 2\theta = \frac{1}{2}\sec\theta

Substitute into the equation:

\tan^2\theta + \frac{7}{2}\sec\theta = 8

Use $\tan^2\theta = \sec^2\theta - 1$ :

\sec^2\theta - 1 + \frac{7}{2}\sec\theta = 8

\sec^2\theta + \frac{7}{2}\sec\theta - 9 = 0

Multiply through by 2:

2\sec^2\theta + 7\sec\theta - 18 = 0

Let $u = \sec\theta$ . Solve $2u^2 + 7u - 18 = 0$ using the quadratic formula:

u = \frac{-7 \pm \sqrt{7^2 - 4(2)(-18)}}{2(2)} = \frac{-7 \pm \sqrt{49 + 144}}{4} = \frac{-7 \pm \sqrt{193}}{4}

Since $\sqrt{193} \approx 13.892$ :

u = \frac{-7 + 13.892}{4} \approx 1.723 \quad \text{or} \quad u = \frac{-7 - 13.892}{4} \approx -5.223

So $\sec\theta \approx 1.723$ or $\sec\theta \approx -5.223$ . Taking reciprocals:

\cos\theta = \frac{1}{1.723} \approx 0.5804 \quad \text{or} \quad \cos\theta = \frac{1}{-5.223} \approx -0.1915

For $\cos\theta = 0.5804$ : since cosine is even, $\theta = \pm \cos^{-1}(0.5804) = \pm 0.952$ .

For $\cos\theta = -0.1915$ : $\theta = \pm \cos^{-1}(-0.1915) = \pm 1.76$ .

All four solutions lie within $-\pi < \theta < \pi$ .

Answer

\theta = \pm 0.952, \; \pm 1.76

Final answer

θ = ±0.952, ±1.76

Detailed explanation

Walkthrough

We must solve $\tan^2\theta + 7\sin\theta \csc 2\theta = 8$ for $-\pi < \theta < \pi$ .

Step 1 — Use part (a). We proved $2\sin\theta \csc 2\theta = \sec\theta$ , so dividing by 2 gives $\sin\theta \csc 2\theta = \frac{1}{2}\sec\theta$ . This replaces the compound term with a simple multiple of $\sec\theta$ .

Step 2 — Substitute. The equation becomes $\tan^2\theta + \frac{7}{2}\sec\theta = 8$ .

Step 3 — Unify the trig ratios. We have both $\tan\theta$ and $\sec\theta$ . The Pythagorean identity $\tan^2\theta = \sec^2\theta - 1$ lets us express everything in terms of $\sec\theta$ alone. Substituting gives $\sec^2\theta - 1 + \frac{7}{2}\sec\theta = 8$ .

Step 4 — Form a quadratic. Rearrange and multiply by 2: $2\sec^2\theta + 7\sec\theta - 18 = 0$ . This is a standard quadratic in the variable $\sec\theta$ .

Step 5 — Solve the quadratic. Let $u = \sec\theta$ . Using the quadratic formula, $u = \frac{-7 \pm \sqrt{193}}{4}$ , giving $u \approx 1.723$ or $u \approx -5.223$ .

Step 6 — Convert to cosine. Since $\sec\theta = \frac{1}{\cos\theta}$ , we have $\cos\theta = \frac{1}{1.723} \approx 0.5804$ or $\cos\theta = \frac{1}{-5.223} \approx -0.1915$ .

Step 7 — Find all angles in the interval. For $\cos\theta = 0.5804$ , the principal value is $\cos^{-1}(0.5804) \approx 0.952$ . Because cosine is even, $\theta = -0.952$ is also a solution. Similarly, $\cos\theta = -0.1915$ gives $\theta = \pm 1.76$ . All four values lie strictly between $-\pi$ and $\pi$ , so all are valid.

Key Takeaways

Reusing an identity proved in an earlier part to simplify a trigonometric equation.
The Pythagorean identity $\tan^2\theta = \sec^2\theta - 1$ for converting between tangent and secant.
Reducing a trigonometric equation to a quadratic and solving by substitution.
When solving $\cos\theta = k$ on $(-\pi, \pi)$ , both $\theta$ and $-\theta$ must be considered.

Common Mistakes

Forgetting the factor of $\frac{1}{2}$ when converting $\sin\theta\csc 2\theta$ using part (a).
Using the wrong Pythagorean identity, e.g. writing $\tan^2\theta = 1 - \sec^2\theta$ .
Only finding the positive solutions and missing $\theta = -0.952$ and $\theta = -1.76$ .
Not showing the quadratic-formula working, which is required for the method marks.

Things to Be Careful About

The interval is open, $(-\pi, \pi)$ ; check that every solution is strictly inside (here $1.76 < \pi$ , so all four are fine).
When $\sec\theta$ is negative, $\cos\theta$ is negative, giving solutions in the second and third quadrants — these appear as the $\pm$ pairs with magnitude greater than $\pi/2$ .
The mark scheme expects exactly four solutions and no others in the interval.
Rounding: the mark scheme accepts $\pm 0.952, \pm 1.76$ or greater accuracy.

Techniques used

substitute the identity from part (a)apply the Pythagorean identity to unify trigonometric ratiossolve a quadratic equation in secanttake reciprocals to find cosinefind all solutions in the given interval

(c)

Find $\int 8\sin^2 \frac{1}{2}x \csc^2 x \,dx$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Rewrite $\csc^2 x$ as $\frac{1}{\sin^2 x}$ and use the double-angle identity $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ to simplify the integrand to a constant multiple of $\sec^2\frac{x}{2}$ . Then integrate using the standard result for $\sec^2(ax+b)$ .

Working

\csc^2 x = \frac{1}{\sin^2 x}

Using $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ :

\sin^2 x = 4\sin^2\frac{x}{2}\cos^2\frac{x}{2}

Substitute into the integrand:

8\sin^2\frac{x}{2} \csc^2 x = 8\sin^2\frac{x}{2} \cdot \frac{1}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}} = \frac{8}{4\cos^2\frac{x}{2}} = \frac{2}{\cos^2\frac{x}{2}} = 2\sec^2\frac{x}{2}

Now integrate. Using $\int \sec^2(ax+b)\,dx = \frac{1}{a}\tan(ax+b) + c$ with $a = \frac{1}{2}$ :

\int 2\sec^2\frac{x}{2}\,dx = 2 \cdot 2\tan\frac{x}{2} + c = 4\tan\frac{x}{2} + c

Answer

4\tan\frac{x}{2} + c

Final answer

4 tan(x/2) + c

Detailed explanation

Walkthrough

We need to find $\int 8\sin^2\frac{x}{2} \csc^2 x \,dx$ .

Step 1 — Rewrite the cosecant. $\csc^2 x = \frac{1}{\sin^2 x}$ .

Step 2 — Use the double-angle identity. The integrand contains $\sin^2\frac{x}{2}$ , and the denominator will contain $\sin^2 x$ . The identity $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ relates these, giving $\sin^2 x = 4\sin^2\frac{x}{2}\cos^2\frac{x}{2}$ .

Step 3 — Simplify. Substituting, the $\sin^2\frac{x}{2}$ factors cancel:

8\sin^2\frac{x}{2} \cdot \frac{1}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}} = \frac{2}{\cos^2\frac{x}{2}} = 2\sec^2\frac{x}{2}

The integrand is now a constant multiple of $\sec^2$ of a linear function of $x$ .

Step 4 — Integrate. Using $\int \sec^2(ax+b)\,dx = \frac{1}{a}\tan(ax+b) + c$ with $a = \frac{1}{2}$ :

\int 2\sec^2\frac{x}{2}\,dx = 2 \cdot 2\tan\frac{x}{2} + c = 4\tan\frac{x}{2} + c

The factor of 2 inside (from $\frac{1}{a} = \frac{1}{1/2} = 2$ ) is essential — it comes from the chain rule in reverse.

Key Takeaways

Simplifying an integrand with reciprocal trig functions using double-angle identities before integrating.
The standard integral $\int \sec^2(ax+b)\,dx = \frac{1}{a}\tan(ax+b) + c$ .
Recognising that $\frac{1}{\cos^2 u} = \sec^2 u$ .

Common Mistakes

Trying to integrate the product $8\sin^2\frac{x}{2}\csc^2 x$ directly without simplifying.
Incorrectly expanding $\sin^2 x$ : forgetting the factor 4 in $4\sin^2\frac{x}{2}\cos^2\frac{x}{2}$ .
Forgetting the factor $\frac{1}{a}$ when integrating $\sec^2(ax+b)$ , i.e. writing $\tan\frac{x}{2}$ instead of $2\tan\frac{x}{2}$ .

Things to Be Careful About

The constant of integration $+c$ should be included (the mark scheme condones its omission, but it is good practice).
Check the coefficient carefully: the simplification must give exactly $2\sec^2\frac{x}{2}$ , not $4\sec^2\frac{x}{2}$ or another multiple.
When integrating $\sec^2\frac{x}{2}$ , the derivative of $\frac{x}{2}$ is $\frac{1}{2}$ , so the antiderivative picks up a factor of 2.

Techniques used

rewrite cosecant as reciprocal of sineapply the double-angle identity to simplify the integrandintegrate secant squared of a linear function

Mathematics 9709/23 — May/June 2024

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