Question

Evaluate : $\int \frac{dx}{3+sinx}$

• A. $\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{3tanx/2+1}{2\sqrt{2}}\right)+c$
• B. $\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{2tanx/2-1}{2\sqrt{2}}\right)+c$
• C. $\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{2tanx/2-1}{2\sqrt{3}}\right)+c$
• D. $\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{3tanx/2+1}{2\sqrt{3}}\right)+c$

Answer 1

Answer: (A)

$\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{3tanx/2+1}{2\sqrt{2}}\right)+c$

Let $I=\int \frac{1}{3+\sin x}dx$
Substitute $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
$I=\int \frac{dt}{(t^2+1)(\frac{2t}{t^2+1}+3)}=\int \frac{2}{3t^2+2t+3}dt$
$=2\int \frac{dt}{{(\sqrt{3}t+\frac{1}{\sqrt{3}})}^2+\frac{8}{3}}=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3t+1}{2\sqrt{2}}\right)$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{3\tan \frac{x}{2}+1}{2\sqrt{2}}\right)$