Question

If 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

Answer: (A)

Let $I=\int \frac{\cos 8x+1}{\cot 2x-\tan 2x}dx$
Now, $D^r=\cot 2x-\tan 2x=\frac{\cos 2x}{\sin 2x}-\frac{\sin 2x}{\cos 2x}$
$=\frac{{\cos }^{2}2x-{\sin }^{2}2x}{\sin 2x\cos 2x}=\frac{2\cos 4x}{\sin 4x}$
$\therefore I=\int \frac{2{\cos }^{2}4x}{\frac{2\cos 4x}{\sin 4x}}dx$
$=\int \frac{2{\cos }^{2}4x\cdot \sin 4x}{2\cos 4x}dx$
$=\frac{1}{2}\int \sin 8xdx=-\frac{1}{2}\frac{\cos 8x}{8}+k$
$=-\frac{1}{16}\cdot \cos 8x+k$
Now, $-\frac{1}{16}\cdot \cos 8x+k=A\cos 8x+k$
$\Rightarrow A=-\frac{1}{16}$