Question

Find the solution of the integral $I=\int \frac{\sin \left(2x\right)}{{\sin }^4\left(x\right)+\mathrm{co}{s}^4\left(x\right)}dx$

Answer 1

Find the solution

$$\begin{gathered} I=\int \frac{\sin \left(2x\right)}{{\sin }^4\left(x\right)+\mathrm{co}{s}^4\left(x\right)}dx \\ =\int \frac{2\sin \left(x\right)\cos \left(x\right)}{{\sin }^4\left(x\right)+\mathrm{co}{s}^4\left(x\right)}dx\left[\because \sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)\right] \end{gathered}$$

Multiplying numerator and denominator with $se{c}^{4}x$ we get

$$\begin{gathered} I=\int \frac{2\sin \left(x\right)\cos \left(x\right)\left({\sec }^{4}x\right)}{({\sin }^4\left(x\right)+{\cos }^4\left(x\right))\left({\sec }^{4}x\right)}dx \\ =\int \frac{2\tan \left(x\right){\sec }^2\mathrm{x}}{1+{\tan }^4\left(x\right)}dx\left[\because \cos \left(x\right)=\frac{1}{\sec \left(x\right)}\mathrm{and}\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}\right] \end{gathered}$$

Let $\tan \left(x\right)=t$

$\therefore dt={\sec }^2\left(x\right)dx$

$$\begin{gathered} I=\int \frac{2t}{1+t^4}dt \\ {t}^2=z \\ \Rightarrow 2tdt=dz \\ I=\int \frac{dz}{1+z^2} \\ ={\tan }^{-1}z+c[\because \int \frac{1}{1+x^2}dx={\tan }^{-1}x] \\ ={\tan }^{-1}\left(t^2\right)+c[\because t^2=z] \\ ={\tan }^{-1}\left({\tan }^2\left(x\right)\right)+c[\because t=\tan \left(x\right)] \end{gathered}$$

Hence, the integral $I={\tan }^{-1}\left({\tan }^2\left(x\right)\right)+c$