Question

\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int \frac{\text{ sin }\left( \text{ 2x }\right)}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[ = \int \frac{2 \sin x \cos x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\text{Dividing numerator and denominator by} \cos^4 x\]
\[ \Rightarrow I = \int \frac{\left( \frac{2 \sin x \cos x}{\cos^4 x} \right)}{\tan^4 x + 1}\text{ dx }\]
\[ = \int \frac{2 \tan x . \sec^2 x}{\left( \tan^2 x \right)^2 + 1} \text{ dx }\]
\[\text{ Let tan}^2 x = t\]
\[ \Rightarrow 2 \tan x \sec^2 x \text  { dx } = dt\]
\[ = \int \frac{dt}{t^2 + 1}\]
\[ \therefore I = \tan^{- 1} \left( t \right) + C\]
\[ = \tan^{- 1} \left( \tan^2 x \right) + C\]