Question ID: #1169
The integral $\int_{0}^{\pi}\frac{8xdx}{4\cos^{2}x+\sin^{2}x}$ is equal to
- (1) $2\pi^{2}$
- (2) $4\pi^{2}$
- (3) $\pi^{2}$
- (4) $\frac{3\pi^{2}}{2}$
Solution:
$$I = \int_{0}^{\pi}\frac{8xdx}{4\cos^{2}x+\sin^{2}x} \quad \text{— (i)}$$
$$I = \int_{0}^{\pi}\frac{8(\pi-x)dx}{4\cos^{2}(\pi-x)+\sin^{2}(\pi-x)}$$
$$I = \int_{0}^{\pi}\frac{8(\pi-x)dx}{4\cos^{2}x+\sin^{2}x} \quad \text{— (ii)}$$
$$(i) + (ii) \Rightarrow 2I = \int_{0}^{\pi}\frac{8\pi dx}{4\cos^{2}x+\sin^{2}x}$$
$$2I = 8\pi \times 2 \int_{0}^{\pi/2}\frac{dx}{4\cos^{2}x+\sin^{2}x}$$
$$I = 8\pi \int_{0}^{\pi/2}\frac{\sec^{2}x}{4+\tan^{2}x}dx$$
$$\text{Let } \tan x = t \Rightarrow \sec^2 x dx = dt$$
$$x \to 0 \Rightarrow t \to 0$$
$$x \to \frac{\pi}{2} \Rightarrow t \to \infty$$
$$I = 8\pi \int_{0}^{\infty}\frac{dt}{4+t^{2}}$$
$$I = 8\pi \left[ \frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right) \right]_{0}^{\infty}$$
$$I = 4\pi \left( \frac{\pi}{2} – 0 \right)$$
$$I = 2\pi^{2}$$
Ans. (1)
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