Question

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

• A. $\cot x+{\cot }^{2}x+c$
• B. $2({\sin }^{5}x+{\cos }^{5}x)+c$
• C. $\frac{-\cot 2x}{2}-\frac{3}{4}x+c$
• D. $\frac{\cot 2x}{2}+\frac{3}{4}x+c$

Answer 1

The correct option is C $\frac{-\cot 2x}{2}-\frac{3}{4}x+c$

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

$\Rightarrow \int \frac{{({\sin }^{2}x+{\cos }^{2}x)}^3-3{\sin }^{2}x\times {\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)}{4{\sin }^{2}x\times {\cos }^{2}x}dx$

$\Rightarrow \int \frac{dx}{{\sin }^{2}2x}-\frac{3}{4}\int dx$

$\Rightarrow \int {cosec}^{2}2xdx-\frac{3}{4}x+c$

$\Rightarrow \frac{-\cot 2x}{2}-\frac{3}{4}x+c$