Evaluate the following integrals:

Solution:

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

$d t=2 \sin x \cos x d x$

we know $\sin 2 x=2 \sin 2 x \cos 2 x$

therefore, $d t=\sin 2 x d x$

$\int \frac{\sin 2 x}{\sqrt{\sin ^{4} x+4 \sin ^{2} x-2}} d x=\int \frac{d t}{\sqrt{t^{2}+4 t-2}}$

Add and subtract $2^{2}$ in denominator

$=\int \frac{d t}{\sqrt{t^{2}+4 t-2}}=\int \frac{d t}{\sqrt{t^{2}+2 \times 2 t+2^{2}-2^{2}-2}}$

Let $\mathrm{t}+2=\mathrm{u}$

$\mathrm{dt}=\mathrm{du}$

$\left.\left.=\int \mathrm{dt} / \sqrt{(}(\mathrm{t}+2)^{2}-6\right)=\int \mathrm{dt} / \sqrt{(} \mathrm{u}^{2}-6\right)$

Since, $\int \frac{1}{\sqrt{\left(x^{2}-a^{2}\right)}} d x=\log \left[x+\sqrt{\left.\left(x^{2}-a^{2}\right)\right]+c}\right.$

$\left.=\int \mathrm{dt} / \sqrt{(} \mathrm{u}^{2}-6\right)=\log \left[\mathrm{u}+\sqrt{\mathrm{u}^{2}-6}+\mathrm{c}\right.$

$=\log \left[t+2+\sqrt{(t+2)^{2}-6}+c\right.$

$=\log \left[t+2+\sqrt{(t+2)^{2}-6}+c=\log \left[\sin ^{2} x+2+\right.\right.$

$\sqrt{\left(\sin ^{2} x+2\right)^{2}-6}+c$