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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Evaluate: $\int\limits^{\pi}_{0} \frac{\text{x \sin x}}{1 + 3\cos^{2}\text{x}}\text{dx}$
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Answer

$\text{Let I} =\int\limits^{\pi}_{0}\frac{\text{x}\sin\text{x}}{1 + 3\cos^{2} \text{x}}\text{dx} $ $\dots\dots\text{(i)}$
$\text{I} = \int\limits^{\pi}_{0}\frac{(\pi - \text{x)}\sin (\pi- \text{x)}}{ 1 + 3\cos^{2}(\pi - \text{x)}} \text{dx}$
$= \int\limits^{\pi}_{0}\frac{\pi \sin \text{x}}{1 + 3\cos^{2}\text{x}}\text{dx} - \int\limits^\pi_0\frac{\text{x} \sin \text{x}}{1 + 3\cos^{2} \text{x}}\text{dx}$
Adding (i) & (ii), we have
$2\text{I} = \int\limits^{\pi}_{0}\frac{\pi\sin\text{x}}{1 + 3\cos^{2} \text{x}}\text{dx}$
$\text{Put}\cos\text{x = t}$
$-\sin\text{x}\text{dx = dt, when x = 0} \Rightarrow \text{t} = 1, \text{for x} = \pi \Rightarrow \text{t = -1}$
$\text{2I} = -\pi\int\limits^{-1}_{1} \frac{\text{dt}}{1 + 3\text{t}^{2}}$
$= \frac{\pi}{3}\int\limits^{1}_{-1}\frac{\text{dt}}{\bigg(\frac{1}{\sqrt{3}}\bigg)^{2} + \text{(t)}^{2}}$
$=\frac{\pi}{3}\times\sqrt{3}\bigg[\tan^{-1}(\sqrt{3\text{t}}\bigg]^{1}_{-1}$
$ = \frac{\sqrt{3\pi}}{3}[\tan^{-1}\sqrt{3} - (\tan^{-1}\sqrt{3})]$
$\text{I} = \frac{\sqrt{3\pi}}{3}.\frac{\pi}{3} = \frac{\sqrt{3\pi^{2}}}{9}$

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