Question

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

• A. $-\frac{1}{16}$
• B. $\frac{1}{16}$
• C. $\frac{1}{8}$
• D. $\frac{1}{9}$

Answer 1

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

$\begin{array}{c}-\int \frac{\cos 8x+1}{\tan 2x-\cot 2x}\\ -\int \frac{\cos 8x+1}{\frac{S_{m}2x}{\cos 2x}-\frac{\cos 2x}{\sin 2x}}dx\\ =-\int \frac{2\sin x.\cos x.\left(\cos 8x+1\right)}{2\left({\sin }^{2}x-{\cos }^{2}2x\right)}dx\\ =\int \frac{\sin 4x\cdot \left(\cos 8x+1\right)}{-2\cos 4x}dx\\ =\frac{1}{2}\int \tan 4x\left[\cos 8x+1\right]dx\\ 4x=t\\ dx=\frac{1}{4}dt\\ =\frac{1}{2\times 4}\int \tan t\left(\cos 2t+1\right)dt\\ =\frac{1}{8}\int \frac{\sin t}{x\cos t}\left(2{\cos }^{2}t-1+1\right)dt\\ =\frac{1}{8}\int \sin 2tdt\\ =\frac{1}{8}\left[\frac{-\cos 2t}{2}\right]+c\\ =\frac{-1}{16}\cos 2t+k=A\cos 8x+k\\ =\frac{-1}{16}\cos 8x+k\\ A=\frac{-1}{16}\end{array}$

Hence, this is the answer.