Integration by substitution

Integration by substitution is the chain rule used backwards¹. We check it with $\int \dfrac{2x}{1 + x^2}\,dx$ .

Replacing the denominator

Putting $t = 1 + x^2$ gives $\dfrac{dt}{dx} = 2x$ , that is $dt = 2x\,dx$ . The $2x\,dx$ in the numerator turns straight into $dt$ .

\int \frac{2x}{1 + x^2}\,dx = \int \frac{dt}{t} = \log|t| = \log\left( 1 + x^2 \right) + C

Differentiating to check, the chain rule gives $\left\{ \log\left( 1 + x^2 \right) \right\}' = \dfrac{1}{1 + x^2} \cdot 2x$ , returning the original.

In a definite integral the limits change too

As $x$ runs from $0$ to $1$ , the variable $t$ runs from $1$ to $2$ .

\int_0^1 \frac{2x}{1 + x^2}\,dx = \int_1^2 \frac{dt}{t} = \log 2 \approx 0.693

Once the limits are converted to $t$ , there is no need to return to $x$ .

The forms it fits

The forms that work are limited: the integrand must look like $g(h(x)) \cdot h'(x)$ , with the derivative of the inner function multiplied in.

Integral	Substitution	Result
$\int \dfrac{2x}{1+x^2}dx$	$t = 1+x^2$	$\log(1+x^2)$
$\int \sin^2 x \cos x\,dx$	$t = \sin x$	$\dfrac{\sin^3 x}{3}$
$\int 2x\cos(x^2)dx$	$t = x^2$	$\sin(x^2)$

When the coefficient does not match, adjust with a constant. For $\int \dfrac{x}{1 + x^2}\,dx$ the answer is $\dfrac{1}{2}\log\left( 1 + x^2 \right)$ : the missing factor of $2$ simply comes outside.

Substituting the other way round

The substitution can also be made in the opposite direction. In $\int_0^1 \sqrt{1 - x^2}\,dx$ , putting $x = \sin\theta$ gives $dx = \cos\theta\,d\theta$ .

\int_0^{\pi/2} \cos^2\theta\,d\theta = \frac{\pi}{4}

That is the area of a quarter circle. This pattern, killing a root with a trigonometric function, recurs throughout integrals involving circles and ellipses.

The humped curve on the graph is $y = \dfrac{2x}{1 + x^2}$ , the ever-rising curve is the antiderivative $y = \log\left( 1 + x^2 \right)$ , and the large dots are $(1, 1)$ and $(1, \log 2)$ .

Integration by substitution, Wikipedia