The integral 0 8 x d x 4 cos 2 x + sin 2 x is equal to [2025]

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Solution

Q.
The integral $\int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ is equal to [2025]

1 $π^{2}$

2 $4 π^{2}$

3 $\frac{3 π^{2}}{2}$

4 $2 π^{2}$

Ans.
(4)

Let $I = \int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}$ ... (i)

$\Rightarrow I = \int_{0}^{π} \frac{8 (π - x) d x}{4 \cos^{2} (π - x) + \sin^{2} (π - x)}$

$\Rightarrow I = \int_{0}^{π} \frac{(8 π - 8 x) d x}{4 \cos^{2} x + \sin^{2} x} d x$ ... (ii)

Adding (i) and (ii), we get

$2 I = 8 π \int_{0}^{π} \frac{d x}{4 \cos^{2} x + \sin^{2} x}$

$\Rightarrow 2 I = 8 π \times 2 \int_{0}^{π} \frac{\sec^{2} x}{4 + \tan^{2 x}} d x$

Put $\tan x = t \Rightarrow \sec^{2} xdx = dt$

$∴ I = 8 π \int_{0}^{\infty} \frac{d t}{4 + t^{2}} = 8 π \times \frac{1}{2} {[\tan^{- 1} \frac{t}{2}]}_{0}^{\infty}$

$= 4 π \times \frac{π}{2} = 2 π^{2}$ .

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