Further Integration (OCR A Level Maths A: Pure): Exam Questions

Find

$\int \frac{4x}{2x^2+5}\mathrm{d}x$

Show that

${\int }_2^{6}4{\left(4x+1\right)}^{\frac{1}{2}}dx=\frac{196}{3}$

Find

$\int -15\sin \left(5x-2\right)\mathrm{d}x$

Find

$\int -3x\sin x\mathrm{d}x$

Use the substitution $u=x+4$ to show that

${\int }_1^2\frac{x}{x+4}\mathrm{d}x=1+4\ln \left(\frac{5}{6}\right)$

Find

$\int \frac{3x^2+2}{2x^3+4x}\mathrm{d}x$

Find

$\int 2\sin x\cos x\mathrm{d}x$

Use algebraic integration to find the exact value of

${\int }_1^3{\left(4x+1\right)}^5\mathrm{d}x$

Show that

$\frac{11}{(2x-3)(x+4)}$

can be written in the form

$\frac{A}{2x-3}+\frac{B}{x+4}$

where $A$ and $B$ are constants to be found.

Hence find

$\int \frac{11}{(2x-3)(x+4)}\mathrm{d}x$

writing your answer in the form

$\ln \left|\mathrm{f}\left(x\right)\right|+c$

where $\mathrm{f}\left(x\right)$ is a function you should find and $c$ is a constant.

The figure below shows a sketch of the curves with equations $y=x^2-3x+4$ and $y=4-x^2+2x$.

The shaded region $R$ is bounded by the two curves.

Use algebraic integration to find the exact area of $R$.

Use the substitution $u=2+\ln x$ to show that

$\int \frac{1}{x{\left(2+\ln x\right)}^3}\mathrm{d}x=\frac{-1}{2{\left(2+\ln x\right)}^2}+c$

where $c$ is a constant.

Find 

$\int \sqrt{1+{\cot }^{2}x}dx$

Use algebraic integration to show that

${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta =\frac{\pi }{8}-\frac{1}{4}$

The figure below shows a sketch of the curve with equation $y=1+2x-\frac{1}{4}{x}^2$ where

• the point  $P(x,y)$ lies on the curve

• the shaded rectangle shown has width $\delta x$ and height $y$

By expressing the series limit as a suitable integral, show that

$\underset{\delta x\to 0}{\lim }\sum _{x=1}^8\left(1+2x-\frac{1}{4}{x}^2\right)\delta x=\frac{329}{12}$

Use algebraic integration to show that

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{2-2{\cos }^2\theta }{\sin 2\theta }\mathrm{d}\theta =\frac{1}{2}\ln 3$

Use algebraic integration to show that

${\int }_0^1\frac{3x{\mathrm{e}}^{-3x^2}}{3-2{\mathrm{e}}^{-3x^2}}\mathrm{d}x=\frac{1}{4}\ln (3-2{\mathrm{e}}^{-3})$

Use algebraic integration to show that

${\int }_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\frac{2\cos x}{1-\cos 2x}\mathrm{d}x=-1$

Use algebraic integration to show that

$\int \left(2x^2-1\right){\mathrm{e}}^x\mathrm{d}x=\left(p{x}^2+qx+r\right){\mathrm{e}}^x+c$

where $p$, $q$ and $r$ are integers to be found and $c$ is a constant.

Show that

$\int \ln x\mathrm{d}x=x\ln x-x+c$

where $c$ is a constant.

Use algebraic integration to find

$\int 5{\cos }^{2}4x\sin 4x\mathrm{d}x$

Use algebraic integration to find

$\int 3x{\left(5x^2+4\right)}^4\mathrm{d}x$

Use algebraic integration to find

$\int 6x^2{\mathrm{e}}^{x^3}\mathrm{d}x$

Use algebraic integration to find

$\int \left(16-32x\right)\sin \left[{\left(4x-2\right)}^2\right]\mathrm{d}x$

Use algebraic integration to find

$\int x^2\sin 3x\mathrm{d}x$

Use algebraic integration to find

$\int \frac{\ln x}{x^3}\mathrm{d}x$

Use the substitution $x=\cos \theta$ to show that

${\int }_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}}\frac{1}{\sqrt{1-x^2}}\mathrm{d}x=\frac{\pi }{12}$

Prove that

$\int \tan kx\mathrm{d}x=\frac{1}{k}\ln \left|\sec kx\right|+c$

where $k$ and $c$ are constants.

Use algebraic integration to show that

${\int }_{\frac{\pi }{18}}^{\frac{\pi }{9}}\frac{{\mathrm{cosec}}^{2}3\theta }{3\cot 3\theta }\mathrm{d}\theta =a\ln 3$

where $a$ is a rational number to be found.

Show that

$\int \frac{8x^2-8x-1}{(4x^2-1)(x-2)}\mathrm{d}x=\ln \left(A|x-2|\sqrt{|4x^2-1|}\right)$

where $A$ is a constant.

Show that

$\int {\mathrm{e}}^x\sin x\mathrm{d}x=\frac{1}{2}{\mathrm{e}}^x\left(\sin x-\cos x\right)+c$

where $c$ is a constant.

The figure below shows a sketch of the curves with equations $y=\sqrt{25-x^2}$     and $y=6x-x^2-5$.

The finite regions bounded by the two curves are shaded.

Show that the $x$-coordinates of the points of intersection are $x=3$, $x=4$ and $x=5$.

[You do not need to solve an equation in $x$.]

Use the substitution $x=5\sin u$ to show that

$\int \sqrt{25-x^2}\mathrm{d}x=\frac{25\arcsin \left(\frac{x}{5}\right)+x\sqrt{25-x^2}}{2}+c$

where $c$ is a constant.

Hence show that the exact area of the shaded regions is

$\frac{25\pi }{4}-4-\frac{25}{2}\left(2\arcsin \left(\frac{4}{5}\right)-\arcsin \left(\frac{3}{5}\right)\right)$