d x 2+ x = (where C is a constant integration.) - Vidyadip

MCQ

$\int \frac{d x}{2+\cos x}=$ (where $C$ is a constant integration.)

A
$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{3}}\right)+C$
B
$\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{3}}\right)+C$
C
$\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{2 \sqrt{3}}\right)+C$
D
$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{2 \sqrt{3}}\right)+C$

✓

Answer

(a) : Let $I=\int \frac{d x}{2+\cos x}$
$=\int \frac{d x}{2+\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}}=\int \frac{\sec ^2 x / 2}{3+\tan ^2 x / 2} d x$
Let $z=\tan \frac{x}{2}$
$\frac{d z}{d x}=\frac{1}{2} \sec ^2 x / 2 ; \sec ^2 \frac{x}{2} d x=2 d z$
So, $I=\int \frac{2 d z}{(\sqrt{3})^2+z^2}=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{z}{\sqrt{3}}\right)+C$
Now substitute $z=\tan \frac{x}{2}$
$I=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{3}}\right)+C$

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