Method of Substitution or Change of Variable

The substitution method mirrors, in reverse, the chain rule used for differentiating a function of a function. If $u$ is a differentiable function of $x$, then $\dfrac{du}{dx}=u'$, i.e. $du=u'\,dx$, so

More generally, for a composite integrand $f[g(x)]\,g'(x)$,

The whole method succeeds or fails on spotting the right substitution — either setting $x=\varphi(u)$ (expressing the old variable in terms of a new one) or $u=g(x)$ (naming a chunk of the integrand as the new variable) so that, after substitution, what remains is an easy integral in $u$. The tell-tale sign that a substitution $u=g(x)$ will work is that the integrand (up to a constant factor) already contains $g'(x)$ multiplying $f(g(x))$.

Typical patterns. If the integrand is (a power of $x$) times $\sqrt{\text{a quadratic in that power}}$, e.g. $2x\sqrt{1+x^2}$, set $u=1+x^2$ so $du=2x\,dx$ and the integral collapses to $\int\sqrt u\,du$. If the integrand is $x\,e^{-x^2}$, set $u=x^2$. If it is $\dfrac{\sin x}{1+\cos x}$, set $u=1+\cos x$, since $du=-\sin x\,dx$ is (up to sign) exactly the numerator. If it is $\dfrac{1}{1+x^2}$, the substitution is trigonometric: put $x=\tan u$, so $dx=\sec^2u\,du$ and $1+x^2=\sec^2u$, and the integral reduces to $\int du=u+c=\tan^{-1}x+c$ — this single substitution is the proof of the standard formula $\int\dfrac{dx}{1+x^2}=\tan^{-1}x+c$. If the integrand is a product like $x(a-x)^8$, set $u=a-x$ (so $x=a-u$) to turn it into a difference of two pure powers of $u$. …