Question

Evaluate:$\int \frac{dx}{3+\sin x}$

• A. $I=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3\tan \frac{x}{2}-1}{2\sqrt{2}}\right)+C$
• B. $I=-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3\tan \frac{x}{2}+1}{\sqrt{2}}\right)+C$
• C. $I=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3\tan \frac{x}{2}+1}{2\sqrt{2}}\right)+C$
• D. None of these

Answer 1

The correct option is C $I=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3\tan \frac{x}{2}+1}{2\sqrt{2}}\right)+C$

Consider the given integral.

$I=\int \frac{1}{3+\sin x}dx$

$I=\int \frac{1}{3+\left(\frac{2{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}dx$

$I=\int \frac{(1+{\tan }^2\frac{x}{2})}{3+3{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}}dx$

$I=\int \frac{{\sec }^2\frac{x}{2}}{3{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}+3}dx$

Let $t=\tan \frac{x}{2}$

$\frac{dt}{dx}={\sec }^2\frac{x}{2}\left(\frac{1}{2}\right)$

$2dt={\sec }^2\frac{x}{2}dx$

Therefore,

$I=2\int \frac{1}{3t^2+2t+3}dt$

$I=\frac{2}{3}\int \frac{1}{t^2+\frac{2t}{3}+1}dt$

$I=\frac{2}{3}\int \frac{1}{t^2+\frac{2t}{3}+\frac{1}{9}-\frac{1}{9}+1}dt$

$I=\frac{2}{3}\int \frac{1}{{(t+\frac{1}{3})}^2+{\left(\frac{2\sqrt{2}}{3}\right)}^2}dt$

We know that

$\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$

Therefore,

$I=\frac{2}{3}\left[\frac{1}{\frac{2\sqrt{2}}{3}}{\tan }^{-1}\left(\frac{t+\frac{1}{3}}{\frac{2\sqrt{2}}{3}}\right)\right]+C$

$I=\frac{1}{\sqrt{2}}\left[{\tan }^{-1}\left(\frac{3t+1}{2\sqrt{2}}\right)\right]+C$

On putting the value of $t$, we get

$I=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3\tan \frac{x}{2}+1}{2\sqrt{2}}\right)+C$

Hence, this is the answer.