First slide

Evaluation of definite integrals

Question

If $\int_{0}^{π / 4} \frac{x \sin x}{\cos^{3} x} d x = \frac{π}{4} + A,$ then $A$ is equal to

Easy

A

0
B

$\frac{1}{2}$
C

1
D

$- \frac{1}{2}$

Solution

$\begin{aligned} \int_{0}^{π / 4} \frac{x \sin x}{\cos^{3} x} d x = \int_{0}^{π / 4} x \tan x \sec^{2} x d x \\ = {x \frac{\tan^{2} x}{2}|}_{0}^{π / 4} - \frac{1}{2} \int_{0}^{π / 4} \tan^{2} x d x \\ = \frac{1}{2} \cdot \frac{π}{4} - \frac{1}{2} \int_{0}^{π / 4} (\sec^{2} x - 1) d x \\ = \frac{1}{2} \cdot \frac{π}{4} - \frac{1}{2} [\tan x - x]^{π / 4} \\ = \frac{1}{2} \cdot \frac{π}{4} - \frac{1}{2} [1 - \frac{π}{4}] = \frac{π}{4} - \frac{1}{2} \end{aligned}$

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