First slide

Evaluation of definite integrals

Question

The value of $I = \int_{0}^{π} \frac{x d x}{4 \cos^{2} x + 9 \sin^{2} x}$ is

Easy

A

$π^{2} / 12$
B

$π^{2} / 4$
C

$π^{2} / 6$
D

$π^{2} / 3$

Solution

Using Property 9, we have
$\begin{array}{l} I = \int_{0}^{π} \frac{(π - x) d x}{4 \cos^{2} (π - x) + 9 \sin^{2} (π - x)} \\ = π \int_{0}^{π} \frac{d x}{4 \cos^{2} x + 9 \sin^{2} x} - \int_{0}^{π} \frac{x d x}{4 \cos^{2} x + 9 \sin^{2} x} \\ 2 I = π \int_{0}^{π} \frac{d x}{4 \cos^{2} x + 9 \sin^{2} x} \\ = 2 π \int_{0}^{π / 2} \frac{d x}{4 \cos^{2} x + 9 \sin^{2} x} \end{array}$
$\begin{array}{l} = 2 π \int_{0}^{π / 2} \frac{\sec^{2} x d x}{4 + 9 \tan^{2} x} \\ = 2 π \int_{0}^{\infty} \frac{d t}{4 + 9 t^{2}} (t = \tan x) \\ = \frac{2 π}{9} \int_{0}^{\infty} \frac{d t}{t^{2} + 4 / 9} = {\frac{2 π}{9} \cdot \frac{3}{2} \tan^{- 1} \frac{3}{2} t|}_{0}^{\infty} = \frac{π^{2}}{6} \end{array}$
Hence $I = π^{2} / 12$

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