In Year 1 you could only integrate powers of $x$ (see Indefinite Integration). Every derivative you learned in Exponentials, Logs and Trig now gives an integral, simply by reading it backwards.
| Function $\mathrm{f}(x)$ | $\displaystyle\int \mathrm{f}(x)\,dx$ | Because… |
|---|---|---|
| $x^n \;(n \neq -1)$ | $\dfrac{x^{n+1}}{n+1} + c$ | Year 1 result |
| $e^x$ | $e^x + c$ | $\dfrac{d}{dx}e^x = e^x$ |
| $\dfrac{1}{x}$ | $\ln|x| + c$ | $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$ |
| $\cos x$ | $\sin x + c$ | $\dfrac{d}{dx}\sin x = \cos x$ |
| $\sin x$ | $-\cos x + c$ | $\dfrac{d}{dx}\cos x = -\sin x$ |
| $\sec^2 x$ | $\tan x + c$ | $\dfrac{d}{dx}\tan x = \sec^2 x$ |
Find $\displaystyle\int \left(3e^{2x} + \frac{4}{x} - 5\cos x\right)dx$.
Differentiating $\mathrm{F}(ax+b)$ by the chain rule gives $a\,\mathrm{F}'(ax+b)$ — an extra factor of $a$. So when integrating you must divide by $a$.
| Integrand | Integral |
|---|---|
| $(ax+b)^n,\; n \neq -1$ | $\dfrac{(ax+b)^{n+1}}{a(n+1)} + c$ |
| $e^{ax+b}$ | $\dfrac{1}{a}e^{ax+b} + c$ |
| $\dfrac{1}{ax+b}$ | $\dfrac{1}{a}\ln|ax+b| + c$ |
| $\cos(ax+b)$ | $\dfrac{1}{a}\sin(ax+b) + c$ |
| $\sin(ax+b)$ | $-\dfrac{1}{a}\cos(ax+b) + c$ |
| $\sec^2(ax+b)$ | $\dfrac{1}{a}\tan(ax+b) + c$ |
Find $\displaystyle\int (2x+1)^5\,dx$.
Find $\displaystyle\int \frac{1}{3x-2}\,dx$, and evaluate $\displaystyle\int_2^5 \frac{3}{x-1}\,dx$ exactly.
Find the exact value of $\displaystyle\int_0^{\pi/6} \sin 3x\,dx$.
Find $\displaystyle\int \frac{x^2}{x^3 + 1}\,dx$ and $\displaystyle\int \tan x\,dx$.
Show that $\displaystyle\int_0^1 \frac{x}{x^2+1}\,dx = \tfrac{1}{2}\ln 2$.
This is the chain rule reversed again: treat the bracket as a single "lump", raise its power by one and divide by the new power. It only works if the derivative of the inside is sitting outside (up to a constant).
Find (a) $\displaystyle\int x(x^2+3)^4\,dx$ (b) $\displaystyle\int \sin x\cos^3 x\,dx$.
There is no standard result for $\sin^2 x$, $\cos^2 x$ or $\tan^2 x$. Rewrite them first using identities from Compound and Double Angles.
$\tan^2 x \equiv \sec^2 x - 1$ · $\sin x\cos x \equiv \tfrac{1}{2}\sin 2x$
Find $\displaystyle\int \sin^2 x\,dx$ and $\displaystyle\int \cos^2 3x\,dx$, and evaluate $\displaystyle\int_0^{\pi/4}\tan^2 x\,dx$ exactly.
Exponential
$\int e^{kx}\,dx = \tfrac{1}{k}e^{kx} + c$
Reciprocal
$\int \tfrac{1}{x}\,dx = \ln|x| + c$
Trig
$\int\cos kx = \tfrac{1}{k}\sin kx$; $\int \sin kx = -\tfrac{1}{k}\cos kx$
sec²
$\int \sec^2 kx\,dx = \tfrac{1}{k}\tan kx + c$
Linear inside
$\int \mathrm{f}(ax+b)\,dx = \tfrac{1}{a}\mathrm{F}(ax+b) + c$
f′/f
$\int \tfrac{\mathrm{f}'}{\mathrm{f}}\,dx = \ln|\mathrm{f}(x)| + c$
f′fⁿ
$\int \mathrm{f}'\mathrm{f}^n\,dx = \tfrac{\mathrm{f}^{n+1}}{n+1} + c$
sin² and cos²
Use $\tfrac{1}{2}(1 \mp \cos 2x)$ first.
tan²
Use $\sec^2 x - 1$, giving $\tan x - x$.
Always
Radians; $+\,c$; check by differentiating.
Find $\displaystyle\int \left(4e^x - \frac{3}{x} + 2\sin x\right)dx$.
▶ Show solution
$\int 4e^x\,dx = 4e^x$, $\int -\tfrac{3}{x}\,dx = -3\ln|x|$, $\int 2\sin x\,dx = -2\cos x$.
$= 4e^x - 3\ln|x| - 2\cos x + c$
Find $\displaystyle\int \left(e^{3x-1} + \cos 4x\right)dx$.
▶ Show solution
Both have a linear inside, so divide by its coefficient.
$= \tfrac{1}{3}e^{3x-1} + \tfrac{1}{4}\sin 4x + c$
Find the exact value of $\displaystyle\int_0^{\pi/8}\sec^2 2x\,dx$.
▶ Show solution
$\Big[\tfrac{1}{2}\tan 2x\Big]_0^{\pi/8} = \tfrac{1}{2}\tan\tfrac{\pi}{4} - \tfrac{1}{2}\tan 0 = \tfrac{1}{2}(1) - 0 = \tfrac{1}{2}$
Show that $\displaystyle\int_1^3 \frac{2}{2x+1}\,dx = \ln\tfrac{7}{3}$.
▶ Show solution
The numerator $2$ is exactly the derivative of $2x+1$, so $\displaystyle\int \frac{2}{2x+1}\,dx = \ln|2x+1|$.
$\Big[\ln(2x+1)\Big]_1^3 = \ln 7 - \ln 3 = \ln\tfrac{7}{3}$ ✓
Find $\displaystyle\int \frac{x^2}{x^3+1}\,dx$ for $x > -1$, then use the same idea to find $\displaystyle\int \frac{6x+3}{x^2+x+5}\,dx$.
▶ Show solution
$\dfrac{d}{dx}(x^3+1) = 3x^2$, so $\displaystyle\int \frac{x^2}{x^3+1}\,dx = \tfrac{1}{3}\ln(x^3+1) + c$.
$\dfrac{d}{dx}(x^2+x+5) = 2x+1$, and $6x + 3 = 3(2x+1)$.
$\displaystyle\int \frac{6x+3}{x^2+x+5}\,dx = 3\ln(x^2+x+5) + c$ (no modulus needed: $x^2+x+5 > 0$ since its discriminant is $1 - 20 < 0$).
By writing $\cot x = \dfrac{\cos x}{\sin x}$, show that $\displaystyle\int_{\pi/6}^{\pi/2}\cot x\,dx = \ln 2$.
▶ Show solution
$\dfrac{d}{dx}\sin x = \cos x$, so $\displaystyle\int \cot x\,dx = \ln|\sin x| + c$.
$\Big[\ln\sin x\Big]_{\pi/6}^{\pi/2} = \ln 1 - \ln\tfrac{1}{2} = 0 + \ln 2 = \ln 2$ ✓
Find $\displaystyle\int x^2(x^3-1)^5\,dx$.
▶ Show solution
The inside $x^3 - 1$ has derivative $3x^2$; we have $x^2$.
Guess $(x^3-1)^6$: derivative $6(x^3-1)^5 \times 3x^2 = 18x^2(x^3-1)^5$.
So $\displaystyle\int x^2(x^3-1)^5\,dx = \frac{(x^3-1)^6}{18} + c$.
In this question you must show detailed reasoning. Find the exact value of $\displaystyle\int_0^{\pi/2}\cos^2 x\,dx$.
▶ Show solution
$\cos^2 x = \tfrac{1}{2}(1 + \cos 2x)$
$\displaystyle\int_0^{\pi/2}\tfrac{1}{2}(1+\cos 2x)\,dx = \Big[\tfrac{1}{2}x + \tfrac{1}{4}\sin 2x\Big]_0^{\pi/2}$
$= \left(\tfrac{\pi}{4} + \tfrac{1}{4}\sin\pi\right) - 0 = \tfrac{\pi}{4}$
Show that $\displaystyle\int_0^{\pi/4}(\sin x + \cos x)^2\,dx = \frac{\pi}{4} + \frac{1}{2}$.
▶ Show solution
$(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x\cos x = 1 + \sin 2x$
$\displaystyle\int_0^{\pi/4}(1 + \sin 2x)\,dx = \Big[x - \tfrac{1}{2}\cos 2x\Big]_0^{\pi/4}$
$= \left(\tfrac{\pi}{4} - \tfrac{1}{2}\cos\tfrac{\pi}{2}\right) - \left(0 - \tfrac{1}{2}\right) = \tfrac{\pi}{4} + \tfrac{1}{2}$ ✓
Water flows into a tank at a rate of $R = 6e^{-0.2t} + 2$ litres per minute, where $t$ is the time in minutes, for $0 \leqslant t \leqslant 10$. The tank is empty at $t = 0$.
(a) Explain why the volume of water in the tank after $T$ minutes is $\displaystyle\int_0^T R\,dt$. (b) Find the exact volume after $5$ minutes and give it to 3 significant figures. (c) Show that the volume after $10$ minutes is $50 - 30e^{-2}$ litres. (d) Describe what happens to the rate of flow if the model were continued for a long time.
▶ Show solution
(a) $R = \dfrac{dV}{dt}$, so the volume is found by integrating the rate. The area under the rate–time graph from $0$ to $T$ is the total amount that has flowed in.
(b) $\displaystyle\int (6e^{-0.2t} + 2)\,dt = \frac{6}{-0.2}e^{-0.2t} + 2t = -30e^{-0.2t} + 2t$
$\Big[-30e^{-0.2t} + 2t\Big]_0^5 = (-30e^{-1} + 10) - (-30) = 40 - 30e^{-1}$
$= 40 - \dfrac{30}{e} \approx 29.0$ litres.
(c) $\Big[-30e^{-0.2t} + 2t\Big]_0^{10} = (-30e^{-2} + 20) + 30 = 50 - 30e^{-2}$ ✓ ($\approx 45.9$ litres)
(d) As $t \to \infty$, $e^{-0.2t} \to 0$, so $R \to 2$: the flow settles to a steady $2$ litres per minute.