Question

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

Answer 1

\[\text{We have}, \]
\[I = \int\frac{\sin x + \cos x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[ = \int\frac{\sin x + \cos x}{\left( \sin^2 x + \cos^2 x \right)^2 - 2 \sin^2 x \cos^2 x} \text{ dx }\]
\[ = \int\frac{\sin x + \cos x}{1 - 2 \sin^2 x \cos^2 x} \text{ dx }\]
\[ = \int\frac{\sin x + \cos x}{1 - \frac{1}{2} \left( 2\sin x \cos x \right)^2} \text{ dx }\]
\[ = \int\frac{\sin x + \cos x}{1 - \frac{1}{2} \sin^2 2x}\text{ dx }\]

\[\text{ Putting  sin x - cos x = t} . . . . . \left( 1 \right)\]
\[ \Rightarrow \left( \sin x - \cos x \right)^2 = t^2 \]
\[ \Rightarrow \sin^2 x + \cos^2 x - 2\sin x \cos x = t^2 \]
\[ \Rightarrow 1 - 2\sin x \cos x = t^2 \]
\[ \Rightarrow \sin 2x = 1 - t^2 \]
\[\text{Differentiating} \left( 1 \right), \text{we get}\]
\[\left( \cos x + \sin x \right)dx = \text{ dt }\]
\[ \therefore I = \int\frac{1}{1 - \frac{1}{2} \left( 1 - t^2 \right)^2}\text{  dt }\]
\[ = \int\frac{2}{2 - \left( 1 - t^2 \right)^2} \text{ dt }\]
\[ = \int\frac{2}{\left( \sqrt{2} \right)^2 - \left( 1 - t^2 \right)^2} \text{ dt }\]
\[ = 2\int\frac{1}{\left( \sqrt{2} + 1 - t^2 \right)\left( \sqrt{2} - 1 + t^2 \right)} \text{ dt}\]

\[= \frac{2}{2\sqrt{2}}\int\left[ \frac{1}{\sqrt{2} + 1 - t^2} + \frac{1}{\sqrt{2} - 1 + t^2} \right]\text{ dt}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{2} + 1 - t^2} \text{ dt}+ \frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{2} - 1 + t^2} \text{ dt}\]
\[ = \frac{1}{\sqrt{2}}\int\frac{1}{\left( \sqrt{\sqrt{2} + 1} \right)^2 - t^2} \text{ dt}+ \frac{1}{\sqrt{2}}\int\frac{1}{\left( \sqrt{\sqrt{2} - 1} \right)^2 + t^2} \text{ dt}\]
\[ = \frac{1}{\sqrt{2}} \times \frac{1}{2\sqrt{\sqrt{2} + 1}}\text{ log }\left| \frac{\sqrt{\sqrt{2} + 1} + t}{\sqrt{\sqrt{2} + 1} - t} \right| + \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{\sqrt{2} + 1}} \tan^{- 1} \frac{t}{\sqrt{\sqrt{2} + 1}} + C\]
\[ = \frac{1}{\sqrt{2}}\left[ \frac{1}{2\sqrt{\sqrt{2} + 1}}\text{ log }\left| \frac{\sqrt{\sqrt{2} + 1} + t}{\sqrt{\sqrt{2} + 1} - t} \right| + \frac{1}{\sqrt{\sqrt{2} + 1}} \tan^{- 1} \frac{t}{\sqrt{\sqrt{2} + 1}} \right] + C, \text{ where t = sin x - cos x}\]