Question

If $\int\limits \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\limits \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 \, 4x \, cos \, 4x \, dx $
$= - \frac{1}{2} \int \sin \, 8x \, dx = \frac{1}{2} \times \frac{\cos \, 8x}{8} + c$
Now, $ \frac{1}{2} \frac{\cos \, 8x}{8} +c = a \, cos \, 8x + c$
$ \therefore \:\: a = \frac{1}{16}$