Integration by substitution

Substitution is the integration counterpart of the chain rule. When an integrand contains a function and (a constant multiple of) its derivative, you can replace the inner function by a new variable and shrink the integral to a standard form.

The idea

If $u = g(x)$, then $du = g'(x)\,dx$. So $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du.$

In practice you choose $u$ to be the "inner" function whose derivative also lives in the integrand. After the substitution, everything in the integrand , including the differential , must be in terms of $u$.

Recognising substitution opportunities

The classical cue: the integrand is a product where one factor is the derivative of another factor (up to a constant). For example, in $\int 2x \cos(x^2)\,dx$, the $2x$ is the derivative of $x^2$, the inner function in $\cos(x^2)$. Set $u = x^2$ and the integral becomes $\int \cos u\,du = \sin u + C = \sin(x^2) + C$.

A second cue: the integrand contains a composition $h(g(x))$ and you suspect that "untangling" $g$ will help. Even when $g'(x)$ is not literally present, sometimes you can manufacture it by multiplying and dividing.

Standard substitutions

Trigonometric substitution

When a radical appears, ordinary $u$-substitution stalls. Trigonometric substitution converts the radical into a single trigonometric function via Pythagorean identities.

For $\sqrt{a^2 - x^2}$, put $x = a\sin\theta$, $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$. Then $a^2 - x^2 = a^2\cos^2\theta$, $\sqrt{a^2-x^2} = a\cos\theta$, and $dx = a\cos\theta\,d\theta$. The integral becomes a trigonometric integral, which you finish using identities, and then convert back to $x$ via a right-triangle picture.

For $\sqrt{a^2+x^2}$, put $x = a\tan\theta$: $a^2 + x^2 = a^2\sec^2\theta$. For $\sqrt{x^2 - a^2}$, put $x = a\sec\theta$.

The $\frac{g'(x)}{g(x)}$ form

Whenever the integrand looks like a derivative divided by the original function, the answer is a logarithm: $\int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)| + C.$ This unifies several common patterns: $\int \tan x\,dx = \int \dfrac{\sin x}{\cos x}\,dx = -\ln|\cos x| + C$, $\int \dfrac{2x}{x^2 + 1}\,dx = \ln(x^2 + 1) + C$, and so on.

Worked examples

Example 1. $\int 2x e^{x^2}\,dx$.

Let $u = x^2$, $du = 2x\,dx$. Integral becomes $\int e^u\,du = e^u + C = e^{x^2} + C$.

Example 2. $\int \sin^4 x \cos x\,dx$.

Let $u = \sin x$, $du = \cos x\,dx$. Integral $= \int u^4\,du = \dfrac{u^5}{5} + C = \dfrac{\sin^5 x}{5} + C$.

Example 3. $\int \dfrac{dx}{x \ln x}$.

Let $u = \ln x$, $du = \dfrac{dx}{x}$. Integral $= \int \dfrac{du}{u} = \ln|\ln x| + C$.

Example 4. $\int \dfrac{x}{1+x^4}\,dx$.

Let $u = x^2$, $du = 2x\,dx$. Integral $= \dfrac{1}{2}\int \dfrac{du}{1+u^2} = \dfrac{1}{2}\tan^{-1}(x^2) + C$.

Example 5. $\int \sqrt{4 - x^2}\,dx$.

Let $x = 2\sin\theta$, $dx = 2\cos\theta\,d\theta$. Then $\sqrt{4-x^2} = 2\cos\theta$. Integral $= \int 2\cos\theta \cdot 2\cos\theta\,d\theta = 4\int \cos^2\theta\,d\theta = 4 \cdot \dfrac{\theta + \sin\theta\cos\theta}{2} = 2\theta + 2\sin\theta\cos\theta + C$. Back-substitute: $\sin\theta = x/2$, $\theta = \sin^{-1}(x/2)$, $\cos\theta = \sqrt{4-x^2}/2$. So the integral equals $2\sin^{-1}(x/2) + \dfrac{x\sqrt{4-x^2}}{2} + C$.

Example 6. $\int \dfrac{\cos x}{\sin^2 x + 4}\,dx$.

Let $u = \sin x$, $du = \cos x\,dx$. Integral $= \int \dfrac{du}{u^2 + 4} = \dfrac{1}{2}\tan^{-1}(u/2) + C = \dfrac{1}{2}\tan^{-1}\left(\dfrac{\sin x}{2}\right) + C$.

Try it yourself

1. $\int 3x^2 \sqrt{x^3 + 1}\,dx$
2. $\int \dfrac{e^{\tan^{-1}x}}{1 + x^2}\,dx$
3. $\int \tan^3 x \sec^2 x\,dx$
4. $\int \dfrac{dx}{x^2 + 9}$
5. $\int \dfrac{\sin(\ln x)}{x}\,dx$
6. $\int \cos x \cos(\sin x)\,dx$
7. $\int \dfrac{dx}{\sqrt{9 - x^2}}$
8. $\int \dfrac{x^2}{\sqrt{1-x^6}}\,dx$
9. $\int \dfrac{dx}{x^2 + 6x + 13}$ (complete the square first)
10. $\int \dfrac{2x+3}{x^2 + 3x + 1}\,dx$
11. $\int \sec^4 x\,dx$ (split off one $\sec^2 x$)
12. $\int \dfrac{dx}{e^x + e^{-x}}$ (multiply numerator and denominator by $e^x$)
13. $\int x\sqrt{1 + x}\,dx$ (let $u = 1 + x$)
14. $\int \dfrac{\sin 2x}{1 + \sin^4 x}\,dx$

Pitfalls and tricks

• The differential must come along. When you set $u = g(x)$, you must also replace $dx$ with $du/g'(x)$.
• Re-substitute at the end of an indefinite integral. The answer must be in terms of the original variable $x$.
• Complete the square before integrating $\dfrac{1}{ax^2 + bx + c}$ , it converts the denominator into $u^2 + k$ or $u^2 - k$, which match standard forms.
• For $\int f(\text{linear})\,dx$, e.g. $\int \sin(3x+2)\,dx$, use $u = 3x+2$ and remember $du = 3\,dx$ , a constant scaling.