Evaluate the following integrals:

Let, $x=\sin ^{2} t$

Differentiating both side with respect to t

$\frac{d x}{d t}=2 \sin t \cos t \Rightarrow d x=2 \sin t \cos t d t$

$y=\int \sqrt{\frac{1-\sin t}{1+\sin t}} 2 \sin t \cos t d t$

$y=\int \sqrt{\frac{(1-\sin t)}{(1+\sin t)} \times \frac{(1-\sin t)}{(1-\sin t)}} 2 \sin t \cos t d t$

$y=2 \int(1-\sin t) \sin t d t$

$y=2 \int \sin t-\frac{1-\cos 2 t}{2} d t$

$y=2\left(-\cos t-\frac{t}{2}+\frac{\sin 2 t}{4}\right)+c$

Again, put $t=\sin \sqrt{x}$

$y=2\left(-\cos \sin \sqrt{x}-\frac{\sin \sqrt{x}}{2}+\frac{\sin (2 \sin \sqrt{x})}{4}\right)+c$

$y=2\left(-\sqrt{1-x}-\frac{\sin \sqrt{x}}{2}+\frac{1}{2} \sqrt{x-x^{2}}\right)+c$