Evaluate the following integrals:

Evaluate $\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{2} x} d x$


$y=\int \frac{\sin 2 x}{\left(\sin ^{2} x\right)^{2}+\left(1-\sin ^{2} x\right)^{2}} d x$

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

Differentiating both side with respect to $x$

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

$y=\int \frac{d t}{t^{2}+(1-t)^{2}}$

$y=\int \frac{d t}{2 t^{2}-2 t+1}$

Try to make perfect square in denominator

$y=\int \frac{d t}{2 t^{2}-2 t+\frac{1}{2}+\frac{1}{2}}$

$y=\int \frac{d t}{(\sqrt{2} t)^{2}-2(\sqrt{2} t)\left(\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}}$

$y=\int \frac{d t}{\left(\sqrt{2} t-\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}}$

Using formula $\int \frac{d t}{t^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{t}{a}$

$y=\frac{1}{\sqrt{2} \times \frac{1}{\sqrt{2}}} \tan ^{-1} \frac{\left(\sqrt{2} t-\frac{1}{\sqrt{2}}\right)}{\frac{1}{\sqrt{2}}}+c$

$y=\sqrt{2} \tan ^{-1}\left(\sqrt{2} t-\frac{1}{\sqrt{2}}\right)+c$

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

$y=\sqrt{2} \tan ^{-1}\left(\sqrt{2} \sin ^{2} x-\frac{1}{\sqrt{2}}\right)+c$


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