Question

$\int \frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x{\cos }^{2}x}dx$

• A. $\sin x-\cot x+C$
• B. $\tan x-\cot x-3x+C$
• C. $\cos x-\cot x+C$
• D. All of the above

Answer 1

The correct option is B $\tan x-\cot x-3x+C$

$\int \frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x{\cos }^{2}x}dx$

$=\int \frac{({\sin }^{2}x+{\cos }^{2x}{)}^3-3{\sin }^{2}x.{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)}{{\sin }^{2}x{\cos }^{2}x}dx$

$=\int \frac{1-3{\sin }^{2}x.{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$

$=\int \frac{1}{{\sin }^{2}x.{\cos }^{2}x}-3dx$

$=\int \frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x.{\cos }^{2}x}-3dx$

$=\int {\sec }^{2}x+\text{cosec}{}^{2}x-3dx$

$=\tan x-\cot x-3x+C$