Integration by substitutionEdexcel A-Level Maths: Revision notes

The chain rule gives $\frac{\mathrm{d}}{\mathrm{d}x}F(g(x))=F'(g(x))\,g'(x)$. Reading it backwards, $\int f(g(x))\,g'(x)\,\mathrm{d}x$ becomes $\int f(u)\,\mathrm{d}u$ with $u=g(x)$. That is integration by substitution. The method: (1) choose $u$; (2) find $\frac{\mathrm{d}u}{\mathrm{d}x}$ and replace $\mathrm{d}x$; (3) rewrite everything in terms of $u$; (4) integrate; (5) return to $x$ (indefinite) or use new limits (definite).

Look for a function whose derivative (up to a constant) is also in the integrand. In $\int x\left(x^2+3\right)^4\mathrm{d}x$ the bracket $x^2+3$ has derivative $2x$, so use $u=x^2+3$: $x\,\mathrm{d}x=\frac12\mathrm{d}u$ and $\int x\left(x^2+3\right)^4\mathrm{d}x=\frac12\int u^4\,\mathrm{d}u=\frac{\left(x^2+3\right)^5}{10}+c.$ Good first choices are an expression inside a power, root, exponential or trigonometric function, or the denominator of a fraction whose numerator is its derivative. Examination questions often give the substitution.

For a definite integral, change the limits to values of $u$ and never return to $x$. Example: $N=\int_0^{\ln2}\frac{e^x}{1+e^x}\mathrm{d}x$ with $u=1+e^x$, $\mathrm{d}u=e^x\mathrm{d}x$. The limits are $x=0\Rightarrow u=2$ and $x=\ln2\Rightarrow u=3$, so $N=\int_2^3\frac1u\,\mathrm{d}u=\big[\ln u\big]_2^3=\ln\frac32.$

Sometimes you must rewrite $x$ in terms of $u$. For $\int_0^4\frac{x}{\sqrt{2x+1}}\mathrm{d}x$ use $u=2x+1$, so $x=\frac{u-1}{2}$ and $\mathrm{d}x=\frac12\mathrm{d}u$. The limits become $1$ and $9$: $\frac14\int_1^9\left(u^{1/2}-u^{-1/2}\right)\mathrm{d}u=\frac14\left[\frac23u^{3/2}-2u^{1/2}\right]_1^9=\frac{10}{3}.$ Using $u=\sqrt x$ gives $x=u^2$, $\mathrm{d}x=2u\,\mathrm{d}u$, which turns $\int_1^4\frac{1}{x+\sqrt x}\mathrm{d}x$ into $\int_1^2\frac{2}{u+1}\mathrm{d}u=2\ln\frac32$.

With $u=\cos x$, $\mathrm{d}u=-\sin x\,\mathrm{d}x$. To integrate $\sin^3x$, write $\sin^3x=\sin x\left(1-\cos^2x\right)$, so $\int\sin^3x\,\mathrm{d}x=-\int\left(1-u^2\right)\mathrm{d}u=-\cos x+\frac{\cos^3x}{3}+c$. Over $\left[0,\frac\pi2\right]$ the new limits are $1$ and $0$, giving $\frac23$. The minus sign from $\mathrm{d}u=-\sin x\,\mathrm{d}x$ can be removed by swapping the limits.