Question

lf the primitive of $\frac{\sin x}{\sqrt{1+\sin x}}$ is $-2\sqrt{f\left(x\right)}-2\log |\tan g\left(x\right)|+c$, then

• A. $f\left(x\right)=1+\sin x$
• B. $g\left(x\right)=\left(3\pi /8\right)-\left(x/4\right)$
• C. $f\left(x\right)=1-\sin x$
• D. $g\left(x\right)=\frac{\pi }{4}+\frac{x}{2}$

Answer 1

Answer: (C), (D)

The correct options are
C $f\left(x\right)=1-\sin x$
D $g\left(x\right)=\frac{\pi }{4}+\frac{x}{2}$

Given, $\int \frac{\sin x}{\sqrt{1+\sin x}}dx=-2\sqrt{f\left(x\right)}+\sqrt{2}\log |\tan g\left(x\right)|+c$ ....(1)
Consider, $\int \frac{\sin x}{\sqrt{1+\sin x}}dx$
$=\int \frac{1+\sin x-1}{\sqrt{1+\sin x}}dx$
$=\int \sqrt{1+\sin x}dx-\int \frac{1}{\sqrt{1+\sin x}}dx$
$=\int (\sin \frac{x}{2}+\cos \frac{x}{2})dx-\int \frac{1}{\sin \frac{x}{2}+\cos \frac{x}{2}}dx$
$=-\frac{\cos \frac{x}{2}}{1/2}+\frac{\sin \frac{x}{2}}{1/2}-\int \frac{1}{\sin \frac{x}{2}+\cos \frac{x}{2}}dx$
Substitute $1=r\cos \theta ,1=r\sin \theta$
$\Rightarrow r=1,\theta =\frac{\pi }{4}$
$=-2(\sin \frac{x}{2}-\cos \frac{x}{2})-\int \frac{1}{r\sin (\frac{x}{2}+\theta )}dx$
$=-2\sqrt{1-\sin x}-\int \frac{1}{\sin (\frac{x}{2}+\frac{\pi }{4})}dx$
$=-2\sqrt{1-\sin x}-\int \csc (\frac{x}{2}+\frac{\pi }{4})dx$
$=-2\sqrt{1-\sin x}-2\log |\tan (\frac{x}{2}+\frac{\pi }{4})|+C$
On comparing with (1), we get
$g\left(x\right)=\frac{x}{2}+\frac{\pi }{4}$ and $f\left(x\right)=\sqrt{1-sinx}$