Question

In the integral
$\int \frac{\cos 8x+1}{\cot 2x-\tan 2x}dx=A\cos 8x+k$, where $k$ is an arbitrary constant, then $A$ is equal to

• A. $-\frac{1}{16}$
• B. $\frac{1}{16}$
• C. $\frac{1}{8}$
• D. $-\frac{1}{8}$

Answer 1

The correct option is A $-\frac{1}{16}$

$\int \frac{\cos 8x+1}{\cot 2x-\tan 2x}dx=\int \frac{2{\cos }^{2}4x-1+1}{\frac{\cos 2x}{\sin 2x}-\frac{\sin 2x}{\cos 2x}}dx=\int \frac{2{\cos }^{2}4x\sin 2x\cos 2x}{{\cos }^{2}2x-{\sin }^{2}2x}dx=\int \frac{{\cos }^{2}4x\sin 4x}{\cos 4x}dx=\int \cos 4x\sin 4xdx=\frac{1}{2}\int \sin 8xdx=\frac{-\cos 8x}{16}+c$

$A=$$-\frac{1}{16}$

so A is the correct option