Integration by substitutionAQA A-Level Maths: Revision notes

The chain rule gives $\frac{d}{dx}F(g(x))=F'(g(x))\,g'(x)$. Reversing it, $\int f(g(x))\,g'(x)\,dx=F(g(x))+c$ where $F'=f$. In integration by substitution a new variable $u=g(x)$ is introduced so the integral becomes $\int f(u)\,du$, which is easier. The key link is $\frac{du}{dx}=g'(x)$, written $du=g'(x)\,dx$. It works when the integrand contains a function and (a multiple of) its derivative, such as $6x(x^2+5)^3$, where $x^2+5$ has derivative $2x$.

1. Choose $u$ (often the inner function of a power, root or exponential, or one given in the question).
2. Differentiate: $\frac{du}{dx}=\dots$, so $dx=\frac{du}{\dots}$.
3. Replace every $x$ and $dx$ by $u$ and $du$. Any leftover $x$ must be written in terms of $u$.
4. Integrate in $u$, then replace $u$ by its expression in $x$ for an indefinite integral. Example: $\int6x(x^2+5)^3dx$ with $u=x^2+5$: $6x\,dx=3\,du$, so $3\int u^3du=\frac34u^4+c=\frac34(x^2+5)^4+c$.

Sometimes $x$ remains after substitution and must be written in terms of $u$. For $K=\int\frac{x}{\sqrt{x+4}}dx$ with $u=x+4$: $du=dx$ and $x=u-4$, so $K=\int\frac{u-4}{u^{1/2}}du=\int\left(u^{\frac12}-4u^{-\frac12}\right)du=\frac23u^{\frac32}-8u^{\frac12}+c.$ Then return to $x$: $K=\frac23(x+4)^{\frac32}-8(x+4)^{\frac12}+c$. Similarly $\int\frac{e^{2x}}{1+e^x}dx$ with $u=1+e^x$ uses $e^x=u-1$ and $du=e^x\,dx$ to give $\int\left(1-\frac1u\right)du$.

For a definite integral change the limits to $u$-values using $u=g(x)$ so you never need to return to $x$. For $\int_0^{\pi/2}\sin x\cos^3x\,dx$ with $u=\cos x$: $x=0\Rightarrow u=1$ and $x=\frac\pi2\Rightarrow u=0$. As $du=-\sin x\,dx$, $J=-\int_1^0u^3du=\int_0^1u^3du=\frac14.$ Reversing the limits cancels the minus sign. If you keep the $x$-limits you must substitute back to $x$ first. Never put $x$-limits on an integral in $u$.

Two patterns are worth recognising: $\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+c$ and $\int f'(x)[f(x)]^n\,dx=\frac{[f(x)]^{n+1}}{n+1}+c$. For example, with $u=\cos x$, $\int\tan x\,dx=-\int\frac{du}{u}=-\ln|\cos x|+c$, so $\int_0^{\pi/3}\tan x\,dx=-\ln\frac12=\ln2$. At A Level only simple cases are tested and the substitution is often supplied; show each step ($du$, new integrand, new limits) because marks are awarded for them. Check by differentiating the answer.