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:

Moderate

A

$π^{2} / 12$
B

$π^{2} / 4$
C

$π^{2} / 6$
D

$π^{2} / 3$

Solution

$I = \int_{0}^{π} \frac{(π - x) d x}{4 \cos^{2} (π - x) + 9 \sin^{2} (π - x)}$
           $= π \int_{0}^{π} \frac{d x}{4 \cos^{2} x + 9 \sin^{2} x} - \int_{0}^{π} \frac{x d x}{4 \cos^{2} x + 9 \sin^{2} x}$
$2 I = π \int_{0}^{π} \frac{d x}{4 \cos^{2} x + 9 \sin^{2} x}$
       $= 2 π \int_{0}^{π / 2} \frac{d x}{4 \cos^{2} x + 9 \sin^{2} x}$
        $= 2 π \int_{0}^{π / 2} \frac{\sec^{2} x d x}{4 + 9 \tan^{2} x} = 2 π \int_{0}^{\infty} \frac{d t}{4 + 9 t^{2}} (t = \tan x)$
        $= \frac{2 π}{9} \int_{0}^{\infty} \frac{d t}{t^{2} + 4 / 9} = \frac{2 π}{9} . \frac{3}{2} \tan^{- 1} {\frac{3}{2} t |}_{0}^{\infty} = \frac{π^{2}}{6}$
$I = π^{2} / 12$

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