Question

Evaluate: $\int \frac{1}{{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x}dx.$

Answer 1

Let $I=\int \frac{1}{{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x}dx$

$=\int \frac{1}{{\sin }^{4}x+2{\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x}dx$

$=\int \frac{1}{({\sin }^{2}x+{\cos }^2{)}^2-\frac{1}{4}{\sin }^2(2x)}$

Multiplying top and bottom by ${\sec }^2\left(2x\right)$ we get

$=\int \frac{{\sec }^2\left(2x\right)}{{\sec }^2\left(2x\right)-\frac{1}{4}{\tan }^2\left(2x\right)}$

Let $u=\tan \left(2x\right)\Rightarrow du=2{\sec }^2\left(2x\right)dx$

$=\int \frac{\frac{1}{2}du}{(1+u^2)-\frac{1}{4}{u}^2}$

$=\int \frac{\frac{1}{2}du}{1+\frac{3}{4}{u}^2}$

$\frac{1}{2}\times \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}}{2}u\right)$

$=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}}{2}\tan \right(2x\left)\right)+C$