# Integration by Parts

## Integration by Parts

If u and v are two functions of x, then $$\int uvdx=u\left( \int udx \right)-\int \left[ \frac{du}{dx}\int udx \right]dx$$ i.e., the integral of the product of two functions = (first function) x (integral of second function) – integral of {diff of first function) x (integral of second function)}

we can choose the first function which comes first in the word ILATE, where

I – Stands for the inverse trigonometric functions (sin⁻¹x, cos⁻¹x, tan⁻¹x, etc).

L – Stands for the logarithmic functions

A – Stands for the algebraic functions

T – Stands for the trigonometric functions

E – Stands for the exponential functions

Example 1: Evaluate ∫x² sinx dx

Solution: Given that,

∫x² sinx dx

$$={{x}^{2}}\left( \int \sin xdx \right)-\int \left[ \frac{d}{dx}\left( {{x}^{2}}. \right)\int \sin xdx \right]dx$$,

= – x² cosx – ∫2x (- cosx) dx

= – x² cosx + 2∫x cosx dx

= – x² cosx + 2[x sinx – ∫sinx dx]

= – x² cosx + 2[x sinx + cosx] + C

Example 2: Evaluate ∫ex (tanx + log sec x) dx

Solution: Given,

∫ex (tanx + log sec x) dx

= ∫ex log sec x dx + ∫ex tan x dx

$$=\left( \log \sec x \right){{e}^{x}}-\int \frac{1}{\sec x}\sec xtanx{{e}^{x}}dx+\int {{e}^{x}}\tan xdx+C$$,

= ex log sec x dx – ∫ex tan x dx + ∫ex tan x dx + C

= ex log (sec x) + C.