Question

If $\int \frac{\cos 8x+1}{\tan 2x-\cot 2x}dx=a\cos 8x+c,$ then $a$ =

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

Answer 1

Answer: (C)

L.H.S = $\int \frac{\cos 8x+1}{\tan 2x-\cot 2x}dx$
$\int \frac{2{\cos }^{2}4x}{\left(\frac{{\sin }^{2}2x-{\cos }^{2}2x}{\sin 2x\cos 2x}\right)}dx$
$=-\int \frac{{\cos }^{2}4x\left(2\sin 2x\cos 2x\right)}{\left({\cos }^{2}2x-{\sin }^{2}2x\right)}dx$
$=-\int \frac{{\cos }^{2}4x\times \sin 4x}{\cos 4x}dx$
$=-\frac{1}{2}\int 2\sin 4xcos4xdx$
$=-\frac{1}{2}\int \sin 8xdx=\frac{1}{2}\times \frac{\cos 8x}{8}+c$
Now, $\frac{1}{2}\frac{\cos 8x}{8}+c=acos8x+c$
$\therefore a=\frac{1}{16}$