Question

$\int\frac{cos\, 4x-1}{cot\, x - tan\, x}dx$ is equal to

• A. $\frac{1}{2} ln \left|sec \,2x\right| - \frac{1}{4} cos^{2} \,2x + c$
• B. $\frac{1}{2} ln \left|sec \,2x\right| -+\frac{1}{4} cos^{2} 2x + c$
• C. $\frac{1}{2} ln \left| cos\, 2x\right| - \frac{1}{4} cos^{2} 2x + c$
• D. $\frac{1}{2} ln \left|cos\, 2x\right| + \frac{1}{4} cos^{2} 2x + c$

Answer 1

Answer: (C)

$I = \int\frac{cos\, 4x - 1}{cot \,x - tan \,x} dx $
$ = \int\frac{-2 \,sin^{2} 2x \left(sin \,x \,cos \,x\right)}{\left(cos^{2} x - sin^{2} x\right)} dx $
$=-\int \frac{sin^{2} \,2x \,sin \,2x}{cos \,2x} x $
$ = \int\frac{\left(cos^{2} 2x -1\right)sin \,2x}{cos \,2x} dx $
Let $t = cos \,2x$ or $dt = -2\, sin \,2x \,dx $.
$\therefore I = \frac{1}{2} \int \frac{(1 - t^2)}{t} dt = \frac{1}{2} ln |t| - \frac{t^2}{4} + C$
$= \frac{1}{2} ln |cos \,2x| - \frac{1}{4} cos^2 \,2x + c$