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)

$-\frac{1}{16}$

Answer 2

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