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Model Paper 7 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 73 Marks
Question
Evaluate: $\int_0^\pi \frac{1}{5+4 \cos x} d x$
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Answer

Let $I=\int_0^\pi \frac{1}{5+4 \cos x} d x$. Then
$I=\int_0^\pi \frac{1}{5+4\left(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)} d x=\int_0^\pi \frac{1+\tan ^2 \frac{x}{2}}{5\left(1+\tan ^2 \frac{x}{2}\right)+4\left(1-\tan ^2 \frac{z}{2}\right)} d x$
$\Rightarrow I=\int_0^\pi \frac{1+\tan ^2 \frac{2}{2}}{9+\tan ^2 \frac{z}{2}} d x=\int_0^\pi \frac{\sec ^2\frac{x}{2}}{9+\tan ^2 \frac{x}{2}} d x$
By using substitution
Let $\tan \frac{x}{2}=t$. Then, $d\left(\tan \frac{x}{2}\right)=d t $
$\Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t$
$ \Rightarrow d x=\frac{2 d t}{\sec ^2 \frac{x}{2}}$
Also, $x=0 \Rightarrow t=\tan 0=0$ and $x=\pi \Rightarrow t=\tan \frac{\pi}{2}=\infty$
$\therefore I=\int_0^{\infty} \frac{\sec ^2 \frac{x}{2}}{9+t^2} \times \frac{2 d t}{\sec ^2 \frac{x}{2}}$
$\Rightarrow I=2 \int_0^{\infty} \frac{d t}{3^2+t^2}=\frac{2}{3}\left(\tan ^{-1} \frac{t}{3}\right)_0^{\infty}
=\frac{2}{3}\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)=\frac{2}{3}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{3}$

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