The integral 0 ( x + 3 ) sin x 1 + 3 cos 2 x d x is equal to [2025]

Question

The integral $\int_{0}^{π} \frac{(x + 3) \sin x}{1 + 3 \cos^{2} x} d x$ is equal to [2025]

Smart Achievers · Accepted Answer

(4)

Let, $I = \int_{0}^{π} \frac{(x + 3) \sin x}{1 + 3 \cos^{2} x} d x$ ... (i)

Replace x to $π$ – x,

$I = \int_{0}^{π} \frac{(π - x + 3) \sin (π - x)}{1 + 3 \cos^{2} (π - x)} d x$

$\Rightarrow I = \int_{0}^{π} \frac{(π + 3) \sin x - x \sin x}{1 + 3 \cos^{2} x} d x$ ... (ii)

Adding equation (i) and (ii), we get

$2 I = \int_{0}^{π} \frac{(x + 6) \sin x}{1 + 3 \cos^{2} x} d x$

Let cos x = t $\Rightarrow$ – sin x dx = dt

$2 I = (π + 6) \int_{1}^{- 1} \frac{- d t}{(1 + 3 t^{2})} = (\frac{π + 6}{3}) \int_{- 1}^{1} \frac{d t}{t^{2} + {(\frac{1}{\sqrt{3}})}^{2}}$

$\Rightarrow 2 I = (\frac{π + 6}{3}) \cdot \frac{1}{(\frac{1}{\sqrt{3}})} \cdot \tan^{- 1} (\frac{t}{1 / \sqrt{3}}) |_{- 1}^{1}$

$\Rightarrow 2 I = (\frac{π + 6}{\sqrt{3}}) [\tan^{- 1} (\sqrt{3}) - \tan^{- 1} (- \sqrt{3})]$

$\Rightarrow 2 I = (\frac{π + 6}{\sqrt{3}}) \cdot (\frac{π}{3} - (- \frac{π}{3})) = (\frac{π + 6}{\sqrt{3}} \cdot \frac{2 π}{3})$

$\Rightarrow I = \frac{π (π + 6)}{3 \sqrt{3}}$ .

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