Integration by Parts

The formula

When a function is a product of two simpler ones, integration by parts breaks it down:

\int u\,dv = uv - \int v\,du

The trick is picking which factor is $u$ (something that gets simpler when differentiated) and which is $dv$ (something easy to integrate).

Choosing u: the ILATE order

Take $u$ to be whichever factor comes first in this list. Functions near the top get simpler when differentiated, and functions near the bottom are easy to integrate.

ILATE: choose u from the type that appears first in the list

Letter	Type	Examples
I	Inverse trigonometric	$\tan^{-1}x$ , $\sin^{-1}x$
L	Logarithmic	$\ln x$
A	Algebraic	$x$ , $x^2$ , $3x + 1$
T	Trigonometric	$\sin x$ , $\cos x$
E	Exponential	$e^x$ , $2^x$

In $\int x\cos x\,dx$ , $x$ is algebraic (A) and $\cos x$ is trigonometric (T). A comes first, so $u = x$ , as in the example below.

Worked example

Find $\int x\cos x\,dx$ .

Solution: Let $u = x$ and $dv = \cos x\,dx$ , so $du = dx$ and $v = \sin x$ :

\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x - (-\cos x) + C

= x\sin x + \cos x + C

Worked example: a lone logarithm

Find $\int \ln x\,dx$ .

Solution: There seems to be only one factor, but write it as $\ln x \cdot 1$ . Take $u = \ln x$ (L) and $dv = dx$ , so $du = \dfrac{1}{x}\,dx$ and $v = x$ :

\int \ln x\,dx = x\ln x - \int x\cdot\dfrac{1}{x}\,dx = x\ln x - x + C

Worked example: applying parts twice

Find $\int x^2e^x\,dx$ .

Solution: Take $u = x^2$ (A) and $dv = e^x\,dx$ :

\int x^2e^x\,dx = x^2e^x - \int 2xe^x\,dx

The new integral still has a product, so apply parts again with $u = 2x$ and $dv = e^x\,dx$ : $\int 2xe^x\,dx = 2xe^x - 2e^x$ . Substituting back:

\int x^2e^x\,dx = x^2e^x - 2xe^x + 2e^x + C = e^x\left(x^2 - 2x + 2\right) + C

Worked example: the cyclic case

Find $I = \int e^x\sin x\,dx$ .

Solution: Take $u = \sin x$ and $dv = e^x\,dx$ :

I = e^x\sin x - \int e^x\cos x\,dx

Apply parts to the new integral, again with the trig function as $u$ ( $u = \cos x$ ):

\int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx = e^x\cos x + I

The original integral $I$ has come back. Substitute and solve for it like an equation:

I = e^x\sin x - e^x\cos x - I \;\Rightarrow\; 2I = e^x(\sin x - \cos x) \;\Rightarrow\; I = \dfrac{e^x}{2}(\sin x - \cos x) + C

Worked example: a definite integral

Evaluate $\displaystyle\int_0^1 xe^x\,dx$ .

Solution: With $u = x$ and $dv = e^x\,dx$ , the antiderivative is $xe^x - e^x$ . So:

\int_0^1 xe^x\,dx = \left[xe^x - e^x\right]_0^1 = (e - e) - (0 - 1) = 1

Common mistakes

Picking $u$ the wrong way round. In $\int xe^x\,dx$ , taking $u = e^x$ and $dv = x\,dx$ gives $\int \dfrac{x^2}{2}e^x\,dx$ , which is harder than the start. Follow ILATE.
Losing the minus sign. The formula is $uv - \int v\,du$ . When applying it twice, the second result goes inside the first one's minus sign.
Switching choices in the cyclic case. If you take the trig function as $u$ the first time, do the same the second time. Switching to $u = e^x$ just undoes the first step.
Forgetting $+ C$ at the end of an indefinite integral.

Where this comes up in exams

The 2025-26 CBSE Class XII syllabus lists integration by substitution, by partial fractions and by parts. JEE Main 2026 lists integration by substitution, by parts and by partial fractions, and JEE Advanced 2026 lists integration by parts alongside substitution and partial fractions.

Checked against CBSE Mathematics curriculum, Classes XI-XII (2025-26), JEE (Main) 2026 syllabus (NTA) and JEE (Advanced) 2026 syllabus. Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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