Evaluate $\int \frac{\sin x}{\sqrt{1+\sin x}} d x$
$\int \frac{\sin x}{\sqrt{1+\sin x}} d x$
We can write above integral as:
$=\int \frac{1+\sin x-1}{\sqrt{1+\sin x}} d x$ (Adding and subtracting 1 in numerator)
$=\int \frac{1+\sin x}{\sqrt{1+\sin x}} d x-\int \frac{1}{\sqrt{1+\sin x}} d x$
$=\int \sqrt{1+\sin x} d x-\int \frac{1}{\sqrt{1+\sin x}} d x$
Consider
$\sqrt{1+\sin x}=\sqrt{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}=\sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}}$
$\left(\because \sin ^{2} x+\cos ^{2} x=1\right.$ and $\left.\sin 2 x=2 \sin x \cdot \cos x\right)$
$\therefore \sqrt{1+\sin x}=\sin \frac{x}{2}+\cos \frac{x}{2} \ldots$ (1)
$\therefore \int \sqrt{1+\sin x} d x-\int \frac{1}{\sqrt{1+\sin x}} d x$
$=\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x-\int \frac{1}{\sin \frac{x}{2}+\cos \frac{x}{2}} d x$
[From (1)]
Considering,
$\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x-\int \frac{1}{\sin \frac{x}{2}+\cos \frac{x}{2}} d x$
$=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}-\int \frac{1}{\frac{2 \tan \frac{x}{4}}{1+\tan ^{2} \frac{x}{4}}+\frac{1-\tan ^{2} \frac{x}{4}}{1+\tan ^{2} \frac{x}{4}}} d x$
$\because \sin \frac{x}{2}=\frac{2 \tan \frac{x}{4}}{1+\tan ^{2} \frac{x}{4}}$ and $\cos \frac{x}{2}=\frac{1-\tan ^{2} \frac{x}{4}}{1+\tan ^{2} \frac{x}{4}}$
$=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}-\int \frac{1+\tan ^{2} \frac{x}{4}}{\left(2 \tan \frac{x}{4}+1-\tan ^{2} \frac{x}{4}\right)+(1-1)} d x$
(Adding and subtracting 1 in denominator)
$=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}+\int \frac{1+\tan ^{2} \frac{x}{4}}{-\left[\left(-2 \tan \frac{x}{4}+1+\tan ^{2} \frac{x}{4}\right)-2\right]} d x$
$=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}-\int \frac{\sec ^{2} \frac{x}{4}}{\left(\tan \frac{x}{4}-1\right)^{2}-2} d x-(2)$
$\because-2 \tan \frac{x}{4}+1+\tan ^{2} \frac{x}{4}=\left(\tan \frac{x}{4}-1\right)^{2}$
Put $\tan \frac{x}{4}-1=u$
$\sec ^{2} \frac{x}{4} d x=4 d u$
Putting values in (2) we get,
$=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}-4 \int \frac{d u}{(u)^{2}-(\sqrt{2})^{2}}$
We know $\int \frac{d u}{(x)^{2}-(a)^{2}}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$
$=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}-4 \frac{1}{2 \sqrt{2}} \log \left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|+C$
Substituting value of $u$ we get,
$=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}-\sqrt{2} \log \left|\frac{\tan \frac{x}{4}-1-\sqrt{2}}{\tan \frac{x}{4}-1+\sqrt{2}}\right|+C$
$\therefore \int \frac{\sin x}{\sqrt{1+\sin x}} d x=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}-\sqrt{2} \log \left|\frac{\tan \frac{x}{4}-1-\sqrt{2}}{\tan \frac{x}{4}-1+\sqrt{2}}\right|+C$