Question

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

• A. $\tan x-\cot x-3x+C$
• B. $\tan x+\cot x-3x+C$
• C. $\tan x-\cot x+3x+C$
• D. ${\tan }^{2}x-\cot x-3x+C$

Answer 1

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

$\int \frac{{\sin }^{6}x+{\cos }^{6}x}{{\sin }^{2}x{\cos }^{2}x}dx\int \frac{({\sin }^{2}x{)}^3+({\cos }^2{)}^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx$
$\{\because a^3+b^3=(a+b\left)\right(a^2+b^2-ab\left)\right\}$
$=\int \frac{({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\cos }^{2}x{\sin }^{2}x)}{{\sin }^{2}x{\cos }^{2}x}dx=\int \frac{\left(1\right)\left(\right({\sin }^{2}x{)}^2+({\cos }^{2}x{)}^2+2{\cos }^{2}x{\sin }^{2}x-3{\cos }^{2}x{\sin }^{2}x)}{{\sin }^{2}x{\cos }^{2}x}dx=\int \frac{({\sin }^{2}x+{\cos }^{2}x{)}^2-3{\cos }^{2}x{\sin }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx=\int \frac{1-3{\cos }^{2}x{\sin }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx=\int \frac{({\sin }^{2}x+{\cos }^{2}x)}{{\sin }^{2}x{\cos }^{2}x}dx-3\int \frac{{\sin }^{2}x{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx$
$\{\because 1={\sin }^{2}x+{\cos }^{2}x\}$
$=\int {\sec }^{2}xdx+\int {\text{cosec}}^{2}xdx-\int 3dx=\tan x-\cot x-3x+C$