Question

Evaluate : $\int \frac{dx}{{\sin }^{6}x+{\cos }^{6}x}$

Answer 1

Consider $I=\int \frac{1}{{\sin }^{6}x+{\cos }^{6}x}$

$I=\int \frac{\frac{1}{{\cos }^{6}x}}{{\tan }^{6}x+1}dx$ (dividing numerator and denominator by ${\cos }^{6}x$)

$=\int \frac{{\sec }^{6}x}{1+{\tan }^{6}x}dx$

$=\int \frac{{\sec }^{4}x{\sec }^{2}x}{1+{\tan }^{6}x}dx=\int \frac{(1+{\tan }^{2}x{)}^2{\sec }^{2}xdx}{1+{\tan }^{6}x}$

Let $\tan x=t$ $\Rightarrow {\sec }^{2}xdx=dt$

$\therefore I=\int \frac{(1+t^2{)}^2}{1+t^6}dt$

$=\int \frac{(1+t^2{)}^{2}dt}{(1{)}^3+(t^2{)}^3}=\int \frac{(1+t^2{)}^{2}dt}{(1+t^2)(1+t^4-t^2)}$ ........ (Applying $a^3+b^3$ formula)

$=\int \frac{1+t^2}{1+t^4-t^2}dt$

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

Take $t-\frac{1}{t}=u$ $\Rightarrow (1+\frac{1}{t^2})dt=du$

$\therefore I=\int \frac{du}{u^2+1}$

$={\tan }^{-1}u={\tan }^{-1}(t-\frac{1}{t})$

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