Evaluate the following integrals:

Assume $a^{2}+b^{2} \sin ^{2} x=t$

$d\left(a^{2}+b^{2} \sin ^{2} x\right)=d t$

$2 b^{2} \cdot \sin x \cdot \cos x \cdot d x=d t$

$(2 \sin x \cdot \cos x=\sin 2 x)$

$\sin 2 x d x=\frac{d t}{b^{2}}$

Put $\mathrm{t}$ and $\mathrm{dt}$ in the given equation we get

$\Rightarrow \frac{1}{\mathrm{~b}^{2}} \int \frac{\mathrm{dt}}{\mathrm{t}}$

$=\frac{1}{\mathrm{~b}^{2}} \ln |\mathrm{t}|+\mathrm{c}$

But $t=a^{2}+b^{2} \sin ^{2} x$

$=\frac{1}{b^{2}} \ln \left|a^{2}+b^{2} \sin ^{2} x\right|+c$