Question

Solve $\int \frac{dx}{2\sin x+\cos x+3}$

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

Answer 1

Answer: (A)

${\tan }^{-1}(\tan \frac{x}{2}+1)+c$

$I=\int \frac{dx}{2\sin x+\cos x+3}$

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

$=\int \frac{1+{\tan }^2\frac{x}{2}dx}{4\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}+3+3{\tan }^2\frac{x}{2}}$

$=\int \frac{{\sec }^2\frac{x}{2}dx}{2{\tan }^2\frac{x}{2}+4\tan \frac{x}{2}+4}$

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

$={\sec }^2\frac{x}{2}.\frac{1}{2}dx=dt$

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

$=\int \frac{2dt}{2t^2+4t+4}=\int \frac{dt}{t^2+2t+2}$

$=\int \frac{dt}{(t+1{)}^2+(1{)}^2}={\tan }^{-1}(t+1)+c$

$={\tan }^{-1}(\tan \frac{x}{2}+1)+c$