Question

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

• A. ${\tan }^{-1}(\tan x+\cot x)$
• B. ${\tan }^{-1}(\tan x-\cot x)$
• C. ${\cot }^{-1}(\tan x-\cot x)$
• D. ${\cot }^{-1}(\tan x+\cot x)$

Answer 1

The correct option is B ${\tan }^{-1}(\tan x-\cot x)$

Let $I=\int \frac{dx}{{\sin }^{6}x+{\cos }^{6}x}=\int \frac{dx}{{\left({\sin }^{2}x\right)}^3{\left({\cos }^{2}x\right)}^3}$
$=\int \frac{dx}{{({\sin }^{2}x+{\cos }^{2}x)}^3-3{\sin }^2{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)}$
$\int \frac{dx}{1-3{\sin }^{2}x{\cos }^{2}x}=\int \frac{{\sec }^{4}x}{{\sec }^{4}x-3{\tan }^{2}x}dx$
$=\int \frac{1+{\tan }^{2}x}{{(1+{\tan }^{2}x)}^2-3{\tan }^{3}x}{\sec }^{2}xdx$
Substitute $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$I=\int \frac{1+t^2}{{(1+t^2)}^2-3t^2}dt=\int \frac{1+t^2}{t^2-t^2+1}dt$
$=\int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt=\int \frac{d(t-\frac{1}{t})}{{(t-\frac{1}{t})}^2+1}$
$={\tan }^{-1}(t+\frac{1}{t})={\tan }^{-1}(\tan x-\cot x)$