Question

Solve: $\int \frac{\sin x}{1+\sin x}dx$

• A. $secx+tanx+x+C$
• B. $secx-tanx+x+C$
• C. $secx+tanx-x+C$
• D. $secx-tanx-x+C$

Answer 1

The correct option is B $secx-tanx+x+C$

$\int \frac{\sin x}{1+\sin x}dx$

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

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

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

$\int \frac{1}{1+\sin x}dx$ $u=\tan \frac{x}{2}$

$1+\sin x=1+u^2+2u$

$=\int \frac{2}{1+u^2+2u}du=\int \frac{2}{(1+u{)}^2}du$

$v=1+u$ $dv=du$

$=\int \frac{2}{v^2}dv=\frac{-2}{v}$

$=x-\left(\frac{-2}{v}\right)$

$=x+2\left(\frac{1}{1+u}\right)$

$=x+2\left(\frac{1}{1+\tan \frac{x}{2}}\right)$

$\int \frac{\sin x}{1+\sin x}dx=\frac{2}{1+\tan \frac{x}{2}}+x+C$.