The integral _ 0 ^ 8 x d x 4 ^ 2 x+ ^ 2 x is equal to - Vidyadip

MCQ

The integral $\int_{0}^{\pi} \frac{8 x d x}{4 \cos ^{2} x+\sin ^{2} x}$ is equal to

A
$2 \pi^{2}$
B
$4 \pi^{2}$
C
$\pi^{2}$
D
$\frac{3 \pi^{2}}{2}$

✓

Answer

A. $2 \pi^{2}$
$I=\int_{0}^{\pi} \frac{8 x d x}{4 \cos ^{2} x+\sin ^{2} x}$
$I=\int_{0}^{\pi} \frac{8(\pi-x) d x}{4 \cos ^{2} x+\sin ^{2} x}$
$2 I=8 \pi \int_{0}^{\pi} \frac{d x}{4 \cos ^{2} x+\sin ^{2} x}$
$2 I=8 \pi \times 2 \int_{0}^{\pi / 2} \frac{\sec ^{2} x}{4+\tan ^{2} x} d x$
$\mathrm{I}=8 \pi \int_{0}^{\infty} \frac{\mathrm{dt}}{4+\mathrm{t}^{2}}=8 \pi \times \frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{t}}{2}\right]_{0}^{\infty}$
$=4 \pi \times \frac{\pi}{2}=2 \pi^{2}$

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