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.