Integration by parts

The product rule of differentiation reads $(uv)' = u'v + uv'$ . Integrate both sides: $uv = \int u'v\,dx + \int uv'\,dx \quad \Longrightarrow \quad \int u\,dv = uv - \int v\,du.$

That last line is integration by parts. It trades one integral for another. The trade is worthwhile only when the new integral is easier than the old.

Statement

If $u$ and $v$ are differentiable functions of $x$ , then $\int u(x)\,v'(x)\,dx = u(x)\,v(x) - \int v(x)\,u'(x)\,dx.$ Equivalently, writing $dv = v'(x)\,dx$ and $du = u'(x)\,dx$ : $\int u\,dv = uv - \int v\,du.$

The art is in choosing $u$ and $dv$ . A poor choice makes the integral harder. A good choice makes it disappear.

LIATE , the choice rule

Among the two factors in your integrand, choose $u$ as whichever class comes first in the list:

Logarithmic ( $\ln x$ )
Inverse trigonometric ( $\sin^{-1}x$ , $\tan^{-1}x$ )
Algebraic ( $x^n$ )
Trigonometric ( $\sin x$ , $\cos x$ )
Exponential ( $e^x$ )

For example, in $\int x\,e^x\,dx$ , the choices are algebraic and exponential. Algebraic comes first in LIATE, so $u = x$ , $dv = e^x\,dx$ . Then $du = dx$ , $v = e^x$ , and $\int x\,e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C.$

Why LIATE? When you differentiate $u$ , you want it to simplify (or at least not blow up). Logs and inverse-trig functions become algebraic on differentiation , they should be $u$ . Trigonometric and exponential factors cycle under differentiation , they make excellent $dv$ .

The single-factor trick

For functions like $\int \ln x\,dx$ or $\int \sin^{-1}x\,dx$ , there is only one factor. Manufacture a second one: set $dv = dx$ , i.e. $v = x$ .

$\int \ln x\,dx$ : $u = \ln x$ , $dv = dx$ . Then $du = \tfrac{1}{x}dx$ , $v = x$ . $\int \ln x\,dx = x \ln x - \int x \cdot \tfrac{1}{x}\,dx = x\ln x - x + C.$

Repeated integration by parts

For $\int x^2 e^x\,dx$ , one application reduces the algebraic factor's degree by one, but you still have an integral. Apply integration by parts again. After two applications, the algebraic factor is gone.

A particularly elegant case occurs when integration by parts twice gives you a multiple of the original integral. Then you solve algebraically. This trick is essential for $\int e^x \sin x\,dx$ and similar.

The $e^x[f(x) + f'(x)]$ shortcut

If you can rewrite an integrand in the form $e^x [f(x) + f'(x)]$ , then $\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C.$

This is integration by parts in disguise: $\int e^x f(x)\,dx = e^x f(x) - \int e^x f'(x)\,dx$ , so $\int e^x f(x)\,dx + \int e^x f'(x)\,dx = e^x f(x)$ . The shortcut saves time on exams.

Worked examples

Example 1. $\int x \cos x\,dx$ .

LIATE: $u = x$ (algebraic), $dv = \cos x\,dx$ . So $du = dx$ , $v = \sin x$ . $\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C.$

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

Single-factor trick: $u = \tan^{-1}x$ , $dv = dx$ . Then $du = \dfrac{dx}{1+x^2}$ , $v = x$ . $\int \tan^{-1}x\,dx = x\tan^{-1}x - \int \frac{x\,dx}{1+x^2} = x\tan^{-1}x - \tfrac{1}{2}\ln(1+x^2) + C.$

Example 3. $\int x^2 e^x\,dx$ .

First: $u = x^2$ , $dv = e^x\,dx$ . So $du = 2x\,dx$ , $v = e^x$ . $\int x^2 e^x\,dx = x^2 e^x - 2\int x e^x\,dx.$ From Example before opening: $\int x e^x\,dx = xe^x - e^x$ . So $\int x^2 e^x\,dx = x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C.$

Example 4. $\int e^x \sin x\,dx$ .

Let $I = \int e^x \sin x\,dx$ . By parts: $u = \sin x$ , $dv = e^x\,dx$ , so $du = \cos x\,dx$ , $v = e^x$ . Then $I = e^x \sin x - \int e^x \cos x\,dx$ . Apply parts again with $u = \cos x$ , $dv = e^x\,dx$ : $\int e^x \cos x\,dx = e^x \cos x + \int e^x \sin x\,dx = e^x \cos x + I$ . Substituting back: $I = e^x \sin x - e^x \cos x - I$ , so $2I = e^x(\sin x - \cos x)$ , giving $I = \dfrac{e^x(\sin x - \cos x)}{2} + C$ .

Example 5. $\int e^x(\sin x + \cos x)\,dx$ .

Here $f(x) = \sin x$ , $f'(x) = \cos x$ . By the shortcut: $e^x \sin x + C$ .

Example 6. $\int e^x \left(\dfrac{1}{x} - \dfrac{1}{x^2}\right)\,dx$ .

Here $f(x) = \dfrac{1}{x}$ , $f'(x) = -\dfrac{1}{x^2}$ . So the integral equals $\dfrac{e^x}{x} + C$ .

Try it yourself

$\int x \sin x\,dx$
$\int x^2 \cos x\,dx$
$\int \sin^{-1}x\,dx$
$\int x \ln x\,dx$
$\int x^2 \ln x\,dx$
$\int e^x \cos x\,dx$
$\int \sec^3 x\,dx$ (use parts with $u = \sec x$ , $dv = \sec^2 x\,dx$ )
$\int e^{2x} \sin 3x\,dx$
$\int x \tan^{-1}x\,dx$
$\int x^3 e^{x^2}\,dx$ (hint: substitute $u = x^2$ first, then parts)
$\int e^x \left(\tan x + \sec^2 x\right)\,dx$ (shortcut)
$\int e^x \dfrac{1 + \sin x}{1 + \cos x}\,dx$ (simplify using half-angle, then shortcut)
$\int (\ln x)^2\,dx$ (apply parts twice)
$\int x \sec^2 x\,dx$

Pitfalls and tricks

LIATE is a guide, not a law. Sometimes the second choice works better. If the new integral is uglier, swap.
Don't keep going forever. If parts is making things worse, stop and reconsider.
Watch for the cycle in $\int e^x \sin x\,dx$ type problems , solve algebraically for the original integral.
The $e^x[f+f']$ pattern is everywhere in JEE. Train your eye to spot it.
For products of a polynomial and $\sin/\cos/e^x$ , parts is the default , use it.