Question

The value of $\int \left(\ln \right(1+\sin x)+x\tan (\frac{\pi }{4}-\frac{x}{2}))dx$ is equal to
(where $C$ is integration constant)

• A. $x\ln (1+\sin x)+C$
• B. $\ln (1+\sin x)+C$
• C. $-x\ln (1+\sin x)+C$
• D. $\ln (1-\sin x)+C$

Answer 1

The correct option is A $x\ln (1+\sin x)+C$

$I=\int \left(\ln \right(1+\sin x)+x\tan (\frac{\pi }{4}-\frac{x}{2}))dx\underset{f\left(x\right)}{\downarrow }\underset{x{f}^{\prime }\left(x\right)}{\downarrow }$

We know
$\frac{d}{dx}\left(\ln \right(1+\sin x\left)\right)=\frac{\cos x}{1+\sin x}=\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{(\cos \frac{x}{2}+\sin \frac{x}{2})}^2}=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}=\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\tan (\frac{\pi }{4}-\frac{x}{2})$
So,
$I=x\ln (1+\sin x)+C[\because \int \left(f\right(x)+x{f}^{\prime }(x\left)\right)dx=xf\left(x\right)+C]$