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Model Paper 3 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 33 Marks
Question
Evaluate $\int_0^{\pi / 2} \frac{x+\sin x}{1+\cos x} d x$
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Answer

$\text { Given } I =\int_0^{\pi / 2} \frac{x+\sin x}{1+\cos x} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{x+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}} d x $
$ {\left[\begin{array}{l}\because \sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \\ \text { and } 1+\cos x=2 \cos ^2 \frac{x}{2}\end{array}\right]}$
$\Rightarrow I=\frac{1}{2} \int_0^{\pi / 2} \underset{I}{x} \sec _{I I}^2 \frac{x}{2} d x+\int_0^{\pi / 2} \tan \frac{x}{2} d x$
$\Rightarrow I=\frac{1}{2}\left\{\left[x \int \sec ^2 \frac{x}{2} d x\right]_0^{\pi / 2}-\int_0^{\pi / 2}\left[\frac{d}{d x}(x) \int\left(\sec ^2 \frac{x}{2} d x\right)\right] d x\right\}+\int_0^{\pi / 2} \tan \frac{x}{2} d x$
$\Rightarrow I=\frac{1}{2}\left\{\left[x \cdot \frac{\tan \frac{x}{2}}{\frac{1}{2}}\right]_0^{\pi / 2}-\int_0^{\pi / 2} \frac{\tan \frac{x}{2}}{\frac{1}{2}} d x\right\}$
$+\int_0^{\pi / 2} \tan \frac{x}{2} d x$
$[$Integration by parts$]$
$=\left[x \cdot \tan \frac{x}{2}\right]_0^{\pi / 2}-\int_0^{\pi / 2} \tan \frac{x}{2} d x+\int_0^{\pi / 2} \tan \frac{x}{2} d x$
$=\frac{\pi}{2} \cdot \tan \frac{\pi}{4}-0$
$\therefore I=\frac{\pi}{2}\left[\because \tan \frac{\pi}{4}=1\right]$

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