Question

$\int \frac{\sin 2x}{a{\cos }^{2}x+b{\sin }^{2}x}dx=$?

Answer 1

We are given that, $I=\int \frac{sin2x}{a\left(co{s}^{2}x\right)+bsi{n}^{3}x}dx$

$I=\int \frac{sin2x}{a(1-si{n}^{2}x)+bsi{n}^{2}x}dx$

$=\int \frac{sin2x}{a+(b-a)si{n}^{2}x}dx$

let $si{n}^{2}x=t$

So, differentiation of above equation

wrt x we get

2 sin x. cos x dx = dt

sin 2x dx = dt

$I=\int \frac{dt}{a+(b-a)t}$

$=\frac{1}{(b-a)}\int \frac{dt}{\frac{a}{b-a}+t}$

$=\frac{1}{b-a}ln|t+\frac{a}{b-a}|+C$

$I=\frac{1}{b-a}ln|si{n}^{2}x+\frac{a}{b-a}|+C$

So, $\int \frac{sin2x}{aco{s}^{2}x+bsi{n}^{2}x}dx=\frac{1}{b-a}ln|si{n}^{2}x+\frac{a}{b-a}|+C$