The integral ${\int }_{0}^{\pi }\frac{(\mathrm{x}+3)\mathrm{sinx}}{1+3{\cos }^{2}\mathrm{x}}\mathrm{dx}$ is equal to : [JEE Main 2025, 7 Apr (Shift 1)]

Question

YESPYQ · Accepted Answer

Correct answer: c) $\frac{\pi }{3\sqrt{3}}(\pi +6)$. Sol: $I={\int }_{0}^{\pi }\frac{(x+3)\sin ⁡x}{1+3{\cos ⁡}^{2}x}dx$ $I={\int }_{0}^{\pi \mathrm{/}2}\frac{(x+3)\sin ⁡x}{1+3{\cos ⁡}^{2}x}\ +\frac{(p−x+3)\sin ⁡(\pi −x)}{1+3{\cos ⁡}^{2}(\pi −x)}dx$ $I={\int }_{0}^{\pi \mathrm{/}2}\frac{(\pi +6)\sin ⁡x}{1+3{\cos ⁡}^{2}x}dx$ $I−(\pi +6){\int }_{0}^{\pi \mathrm{/}2}\frac{\sin ⁡x}{1+3{\cos ⁡}^{2}x}$ $\begin{matrix}I=(\pi +6){\int }_{1}^{0}\frac{−dt}{1+3{t}^{2}}\ (∵\cos ⁡x=t) \ =\frac{\pi +6}{3}{\int }_{0}^{1}\frac{dt}{{(\frac{1}{\sqrt{3}})}^{2}+{t}^{2}} \ =\frac{\pi +6}{3}⋅\sqrt{3}⋅{({\tan ⁡}^{−1}\sqrt{3}t)}_{0}^{1} \ =\frac{\pi +6}{\sqrt{3}}⋅\frac{\pi }{3}\end{matrix}$

The integral \({\int }_{0}^{\pi }\frac{(\mathrm{x}+3)\mathrm{sinx}}{1+3{\cos }^{2}\mathrm{x}}\mathrm{dx}\) is equal to :…

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