Integration by Substitution

When the integrand is not a standard form directly but can be seen as a composite function multiplied by the derivative of its inner function, the method of substitution converts it

into a standard integral in a new variable.

Derivation from the chain rule. Suppose $F$ is an antiderivative of $f$, so $F'=f$, and let

$u=g(x)$ where $g$ is differentiable. By the chain rule,

Reading this equation as a statement about antiderivatives (i.e. integrating both sides),

Writing $u=g(x)$ so that $du=g'(x)\,dx$, the left side becomes exactly $\int f(u)\,du=F(u)+C$ --

the substitution has turned a composite integral in $x$ into a (hopefully standard) integral in

$u$, to be converted back to $x$ at the very end by resubstituting $u=g(x)$.

Practical recipe. (i) Identify a part of the integrand, $u=g(x)$, whose derivative $g'(x)$

also appears (up to a constant multiple) elsewhere in the integrand; (ii) compute $du=g'(x)\,dx$

and rewrite the ENTIRE integral in terms of $u$ only, with no $x$ remaining; (iii) integrate the

resulting standard form in $u$; (iv) substitute $u=g(x)$ back to return to $x$.

Worked illustration. For $\displaystyle\int 2x\cos(x^2)\,dx$: let $u=x^2$, so $du=2x\,dx$ --

exactly the factor $2x\,dx$ present in the integral. The integral becomes $\int\cos u\,du=\sin u+C=\sin(x^2)+C$, matching Example 2.

Substitution with a trigonometric identity. Sometimes the integrand needs an identity applied

FIRST before a substitution becomes visible. For $\int\sin^3x\,dx$, write

$\sin^3x=\sin x\cdot\sin^2x=\sin x\,(1-\cos^2x)$ using $\sin^2x+\cos^2x=1$; now $u=\cos x$ gives …