Integration: past paper questions & worked solutions

First split the fraction into powers:

Integrate term by term. Since $x\gt 0$, use $\ln x$ without absolute-value notation:

Use $\sin^2(3x)=(1-\cos6x)/2$:

The derivative of $(x^2+4)^{5/2}$ is $5x(x^2+4)^{3/2}$. Therefore

For example, differentiating this answer gives

which recovers the integrand. Each indefinite integral has its own arbitrary constant.

The denominator derivative is $6x-4=2(3x-2)$. Hence reverse the logarithmic chain rule:

The log argument is always positive because $3x^2-4x+5=3(x-2/3)^2+11/3$.

Put $u=e^{2x}-1$, so $du=2e^{2x}\,dx$. Then $\int e^{2x}(e^{2x}-1)^{-3}\,dx=(1/2)\int u^{-3}\,du=-1/(4u^2)+C$. The source excludes $x=0$, where the integrand is undefined.

First expand and use the double-angle identity:

Hence

Expand the square first:

Use $\cos^2x=(1+\cos2x)/2$, $\sin^2x=(1-\cos2x)/2$, and $2\sin x\cos x=\sin2x$. The integrand becomes

Integrating with the angle-2 chain factors gives $5x/2+3\sin2x/4+\cos2x+K$.

Apply the cosine addition formula:

Using $\sin^2 A=1-\cos^2 A$ gives the required identity.

From part (a), $\cos^2 3x=(1+\cos6x)/2$. Therefore

Since $\sin(3\pi/4)=\sqrt2/2$ and $\sin(\pi/2)=1$, the result is

Since $d(3x^2+k)/dx=6x$, the numerator is $3/2$ times this derivative. The denominator is positive because $k\gt 0$, so

Apply the limits to (a) and combine the logarithms:

Hence $\ln((75+k)/(12+k))=(2/3)\ln8=\ln4$, so $75+k=4(12+k)$. This gives $27=3k$ and $k=9$, consistent with $k\gt 0$.

Raise the power by one, divide by the new power and by the inner derivative 2:

The derivative of $1+2\cos x$ is $-2\sin x$, so a primitive is $-2\ln(1+2\cos x)$. The log argument is positive throughout $0\le x\le\pi/3$. Hence

Write the integrand as $8(2x-3)^{-3}$. A primitive is $-2(2x-3)^{-2}$, because its derivative is $8(2x-3)^{-3}$. Thus

Since $d(x^2+3)/dx=2x$, reverse the chain rule:

Using the chain rule,

Treat $k$ as a constant when integrating:

The stated inequality therefore implies

Since the exponential is increasing and $k+3\gt 0$,

The roots are $(21\pm\sqrt{697})/2$, so the positive upper bound is $k\lt (21+\sqrt{697})/2=23.700378782\ldots$. With $k\gt 1$ an integer, the greatest possible value is $k=23$.

Multiplying the proposed identity by $x-2$ gives

Thus $A=2$, $B-4=-5$ and $C-2(-1)=8$, giving $B=-1$, $C=6$. Hence $g(x)=2x-1+6/(x-2)$, for $x\ne2$.

Since $x-2\gt 0$ throughout $[4,8]$,

Differentiate the logarithm directly on its domain $\sin3x\ne0$:

This does not assume $\sin3x\gt 0$; the logarithm's argument is its square.

Apply the power and chain rules:

From part (ii)(a), an antiderivative of $x(3x^2-4)^5$ is $(3x^2-4)^6/36$.

Multiply the proposed identity by $(x+3)^2=x^2+6x+9$ and compare coefficients:

The coefficient of $x$ checks: $9B+6C=27-6=21$. The constant equation is $9C+D=-4$, so $D=5$. Hence $f(x)=2x^2+3x-1+5(x+3)^{-2}$.

Integrate each term in the decomposition:

Here $K$ is an arbitrary constant; the domain $x\gt -3$ avoids the denominator zero.

Expand and use $\cos^22x=(1+\cos4x)/2$:

The zeros satisfy $\cos2x=-1/2$. The first two positive values are $x=\pi/3$ and $x=2\pi/3$, so $a=2\pi/3$. The squared curve is nonnegative, so integrate directly for the total shaded area:

Use $\sin(4\pi/3)=-\sqrt3/2$ and $\sin(8\pi/3)=\sqrt3/2$ to obtain $\mathcal A=2\pi-\sqrt3+\sqrt3/4=2\pi-3\sqrt3/4$.

Write the integrand as $12(2x-1)^{-2}$. Reverse the chain rule, dividing by the inner derivative 2:

(a) Since $4x+3=4(x+2)-5$,

(b) A primitive is $4x-5\ln|x+2|$. The absolute value is essential here: $x+2$ is negative throughout $[-8,-5]$.

Start with the angle-addition identity:

Using $\sin2x=2\sin x\cos x$ and $\cos2x=1-2\sin^2x$,

From (a), $\sin^3x=(3\sin x-\sin3x)/4$. Therefore, using algebraic integration,

Multiply the identity by $x^2+4$:

Comparing the cubic, quadratic and linear coefficients gives $A=4$, $B=2$ and $4A+C=3$, hence $C=3-16=-13$.

The $x^2$ coefficients give $B=2$, while the constant terms give $4B+D=8$. Substitute $B=2$: $4(2)+D=8$, so $D=0$.

Using the decomposition, the integrand is $4x+2-13x/(x^2+4)$. Since the derivative of $x^2+4$ is $2x$, an antiderivative is

Therefore

Combine the logarithms: $36-\frac{13}{2}\ln4=36-13\ln2$.

For $y=(4x-7)^{1/2}$,

At $P=(8,5)$, the tangent gradient is $2/5$, so the normal gradient is $-5/2$. The point-gradient equation is

The curve meets the $x$-axis at $x=7/4$. The normal meets it at $x=10$. Split the shaded region at $P$, into a curve area from $7/4$ to 8 and a triangle from 8 to 10:

The triangle has area $\frac12(10-8)\cdot5=5$. Hence the total area is $125/6+5$.

Clear the denominator in the proposed identity:

Matching coefficients gives $A=2$, $3A+B=0$, hence $B=-6$. Then $4A+3B+2C=-4$ gives $C=3$; the constant checks as $4B+3C=-24+9=-15$. Therefore $f(x)=2x-6+3(2x+3)/(x^2+3x+4)$.

The numerator $2x+3$ is the derivative of $x^2+3x+4$. This quadratic is $(x+3/2)^2+7/4\gt 0$, so

Using both limits,

Thus $p=4$ and $q=8$ are integers, as required.

Increase the power to $8$ and divide by both $8$ and the inner derivative $3$:

The denominator is $x^2+x-2$. The first quotient term is $4x$. Subtracting $4x(x^2+x-2)$ leaves $-3x^2+x+14$. The next quotient term is $-3$. Subtracting $-3(x^2+x-2)$ leaves $4x+8$. Thus polynomial division gives

Since $4x+8=4(x+2)$,

Hence $a=4$, $b=-3$, $c=4$.

For $x\gt 1$, an antiderivative is $2x^2-3x+4\ln(x-1)$, so

Thus the exact integral is $126+8\ln2$.

Using a common denominator,

The stated exclusion $x\ne n\pi/2$ ensures both denominators are nonzero.

By part (a), the squared bracket is $\csc^2\theta$. Using $\csc^2\theta=1+\cot^2\theta$ and setting $u=\cot\theta$ gives

On $0\lt \theta\lt \pi/2$, both positive cotangents are permitted: $u=1$ gives $\theta=\pi/4$, while $u=5$ gives $\tan\theta=1/5$, $\theta=0.1973955598\ldots$ radians.

The bracket in part (a) equals $\csc x$. Therefore

Since $3x-1\gt 0$ throughout $3\le x\le42$, an antiderivative is

The factor $1/3$ compensates for the derivative of $3x-1$. Thus

Using $\ln(a^3)=3\ln a$ for $a\gt 0$, the exact value in simplest form is

Multiply the given decomposition by $(x-1)^2$:

Comparing coefficients gives

The coefficient of $x$ also agrees, since $A-2B=2-2(-3)=8$. Hence

Integrating each term,

Therefore, for $x\gt 1$,

where $K$ is an arbitrary constant of integration, distinct from the coefficient $C=4$.

The absolute value allows both intervals on which the integrand is defined.

The largest value of $a-|bx-1|$ occurs when $bx-1=0$. The maximum point is therefore $\boxed{(1/b,a)}$.

At $x=0$, $9=a-1$, so $\boxed{a=10}$. At the intersection $x=3$, the first curve has height $11/11+4=5$. Thus $10-|3b-1|=5$, giving $3b-1=5$ or $-5$. These give $b=2$ or $-4/3$; only the positive integer is allowed, so $\boxed{b=2}$.

For $x\geq3$, the required branch of $C_2$ is $y=10-(2x-1)=11-2x$. Its $x$-intercept is $x=11/2$. The shaded area is the area under $C_1$ from 0 to 3 plus the triangle from 3 to $11/2$:

Hence $\boxed{A=14.4\text{ square units}}$ to 3 significant figures.

Write $y=(16/9)(3x-k)^{-1}$. The chain rule gives

At $x=1$, the given gradient yields

Hence $3-k=\pm2/3$. Both resulting k-values are positive and different from 3.

Since $k\lt 3$, use $k=7/3$. At $x=1$, $y=16/[9(3-7/3)]=8/3$. The tangent gradient is $-12$, so the normal gradient is $1/12$:

With $k=7/3$, the denominator does not vanish on $[1,3]$. Using the reciprocal-linear antiderivative,

Differentiate the velocity:

At a stationary velocity, $e^{t-10}=9e^{-0.75t}$, so $e^{1.75t}=9e^{10}$ and $t=(10+\ln9)/1.75=6.9698426\ldots$ seconds. The second derivative $-e^{t-10}-6.75e^{-0.75t}$ is negative, so this is a maximum. Substitute this unrounded time into v to get $11.88728125\ldots$ metres per second.

Integrating the velocity gives $12t-e^{t-10}+16e^{-0.75t}$. Evaluate between 0 and T:

Rearrange to $12T=116-16e^{-0.75T}+e^{T-10}-e^{-10}$ and divide by 12.

With $T_1=10$, substitution gives $T_2=9.7492587708\ldots$. Continuing with unrounded values gives

Further iteration settles at $9.7293319307\ldots$ seconds. Thus (i) $T_2=9.7493$ and (ii) the race time is $9.7293$ seconds, to four decimal places.

With $u=6x^2+4x-2$ and $v=2x+1$, the quotient rule gives

At $x=2$, $f'(2)=80/25=16/5$, so the normal gradient is $-5/16$. Through $P=(2,6)$,

Multiplying the proposed identity by $2x+1$ gives

Compare coefficients: $2A=6$, $A+2B=4$, $B+D=-2$. Thus $f(x)=3x+\frac12-\frac{5}{2(2x+1)}$.

The curve meets the $x$-axis when $6x^2+4x-2=2(3x-1)(x+1)=0$. The root $x=-1$ is outside $x\gt -1/2$, leaving $x=1/3$. The normal meets the axis at $x=106/5$. Split the shaded region at $P$: a curve area from $1/3$ to 2, then a triangle.

The triangle has area $\frac12(106/5-2)\cdot6=288/5$. Adding gives the exact area of $R$.

Factor the denominator as $(x-4)(x+3)$ and multiply the proposed identity through by it:

The right side expands to $x^4-x^3+(P-12)x^2+(Q-P)x+3Q-12P$. Comparing coefficients gives $P-12=-10$, so $P=2$, and $Q-P=3$, so $Q=5$. The constant check is $3(5)-12(2)=-9$, as required. The identity holds where the original fraction is defined: the stated $x\gt -3$ still excludes $x=4$.

From (a), $g(x)=x^2+2+5/(x-4)$ on the curve domain. Thus $g'(x)=2x-5/(x-4)^2$. At $x=2$, $g(2)=4+2-5/2=7/2$ and $g'(2)=4-5/4=11/4$. The tangent through $(2,7/2)$ is

The source shading is above the x-axis from $x=0$ to $x=2$, so integrate g over those bounds.

The absolute value is essential because $x-4\lt 0$ throughout this interval. Evaluating gives

Thus $a=20/3$ and $b=-5$ in the requested form.