Question

If $\int \frac{cos8x+1}{tan2x-cot2x}dx=acos8x+C$, then

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

Answer 1

The correct option is C $a=\frac{1}{16}$

$\int \frac{cos8x+1}{tan2x-cot2x}dx=\int \frac{2co{s}^{2}4x}{si{n}^{2}2x-co{s}^{2}2x}.sin2xcos2xdx=-\int sin4xcos4xdx=-\frac{1}{2}\int sin8xdx=\frac{1}{16}cos8x+C\therefore a=\frac{1}{16}$