# Evaluate :$\int \large\frac{\sin 2x}{a^2\sin^2x+b^2\cos^2x}$$dx ## 1 Answer Toolbox: • Method of substitution: • Given \int f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t). • Consider I=\int f(x)dx. • Put x=g(t) so that \frac{dx}{dx}=g'(t). • dx=g'(t)dt. • Thus I=\int f(g(t).g'(t))dt. • \int \sqrt{a^2+x^2}dx=\large\frac{x}{2}$$\sqrt{x^2+a^2}+\large\frac{a^2}{2}$$\log \mid x+\sqrt{x^2+a^2}\mid+c Step 1: Let I=\int \large\frac{\sin 2x}{a^2\sin^2 x+b^2\cos^2x}$$dx$
Put $a^2\sin^2x+b^2\cos^2x=t$
Differentiating with respect to $x$
$(2a^2\sin x\cos x-2b^2\sin x\cos x)dx=dt$
$\Rightarrow dx=\large\frac{dt}{2\sin x\cos x(a^2-b^2)}$
Substituting this we get,
$I=\int \large\frac{\sin 2x}{t}.\frac{dt}{2\sin x\cos x(a^2-b^2)}$
But $2\sin x\cos x=\sin 2x$
Step 2:
$I=\int \large\frac{dt}{t(a^2-b^2)}$
On integrating we get,
$\;\;=\large\frac{1}{a^2-b^2}$$\log \mid t\mid+c Substituting for t we get, I=\large\frac{1}{a^2-b^2}$$\log \mid a^2\sin^2 x+b^2\cos ^2 x\mid+c$