Evaluate the following integrals:

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

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

$(-2 a \cos x \cdot \sin x+2 b \sin x \cdot \cos x) d x=d t$

$(b \sin 2 x-a \sin 2 x) d x=d t$

$(b-a) \sin 2 x d x=d t$

$\sin 2 x d x=\frac{d t}{(b-a)}$

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

$\Rightarrow \frac{1}{(b-a)} \int \frac{d t}{t}$

$=\frac{1}{\mathrm{~b}-\mathrm{a}} \ln |\mathrm{t}|+\mathrm{c}$

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

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