Write a value We know that $cos ^{2} x=1-sin ^{2} x$


Write a value of $\int \frac{\sin 2 x}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x$


We know that $\cos ^{2} x=1-\sin ^{2} x$

$\left(a^{2} \sin ^{2} x+b^{2} \cos ^{2} x\right)=a^{2} \sin ^{2} x+b^{2}\left(1-\sin ^{2} x\right)$

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

$y=\int \frac{\sin 2 x}{\left(a^{2}-b^{2}\right)(\sin x)^{2}+b^{2}} d x$

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

Differentiating both sides with respect to $x$

$\frac{d t}{d x}=2 \sin x \cos x$

$=\sin 2 x$

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

$y=\int \frac{d t}{\left(a^{2}-b^{2}\right) t+b^{2}}$

Use formula $\int \frac{1}{c t+d} d t=\frac{\log (c t+d)}{c}$

$y=\frac{\log \left[\left(a^{2}-b^{2}\right) t+b^{2}\right]}{\left(a^{2}-b^{2}\right)}+c$

Again, put $t=\sin ^{2} x$

$y=\frac{\log \left[\left(a^{2}-b^{2}\right)(\sin x)^{2}+b^{2}\right]}{\left(a^{2}-b^{2}\right)}+c$

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