Question

Evaluate: $\int \frac{1}{3+2\sin x+cosx}dx$

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

Answer 1

Answer: (B)

$I={\tan }^{-1}[\tan \left(\frac{x}{2}\right)+1]+c$

Let $I=\int \frac{1}{3+2\sin x+\cos x}dx$

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

$x=2{\tan }^{-1}t$

$dx=\frac{2dt}{(1+t^2)}$ and $\sin x=\frac{2t}{(1+t^2)}$

$\cos x=\frac{(1-t^2)}{1+t^2)}$

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

$=\int \frac{2dt}{3+3t^2+4t+1-t^2}$

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

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

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

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

$={\tan }^{-1}(t+1)+c$

Therefore,

$I={\tan }^{-1}[\tan \left(\frac{x}{2}\right)+1]+c$