Question

$\int \left(\frac{{\cot }^{2}2x-1}{2\cot 2x}-\cos 8x\cot 4x\right)dx=$

• A. $-\frac{\cos 8x}{8}+c$
• B. $\frac{\cos 8x}{8}+c$
• C. $-\frac{\sin 8x}{8}+c$
• D. $\frac{\sin 8x}{8}+c$

Answer 1

Answer: (A)

$-\frac{\cos 8x}{8}+c$

$\frac{{\cot }^{2}2x-1}{2\cot 2x}=\frac{1-{\tan }^{2}2x}{2\tan 2x}=\cot 4x$

$\int (\frac{{\cot }^{2}2x-1}{2\cot 2x}-\cos 8x\cot 4x)dx=\int \left(\cot 4x\right)(1-\cos 8x)dx=\int \left(2{\sin }^{2}4x\right)\left(\cot 4x\right)dx=\int \sin 8xdx=-\frac{\cos 8x}{8}+c$