Q. If $f(x) = \cos{x - \cos^{2}{x + \cos^{3}{x - ...\infty}}}$, then $\int_{}^{}{f(x)dx}$ equals
A) $\tan{\frac{x}{2} + c}$
B) $x + \tan{\frac{x}{2} + c}$
C) $x - \frac{1}{2}\tan{\frac{x}{2} + c}$
D) $x - \tan{\frac{x}{2} + c}$
Explanation
(d)
$\because f(x) = \cos{x - \cos^{2}{x + \cos^{3}{x - ...\infty}}}$
$= \frac{\cos x}{1 + \cos x}$
$\therefore\int_{}^{}{f(x)}\ dx = \int_{}^{}{\frac{1 + \cos x}{1 + \cos x}dx - \int_{}^{}{\frac{1}{1 + \cos x}\ dx}}$
$= \int_{}^{}1\ dx - \frac{1}{2}\ \int_{}^{}{\sec^{2}\frac{x}{2}}\ dx$
$= x - \frac{1}{2}\tan\frac{x}{2}\ .2 + c$
$= x - \tan{\frac{x}{2} + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.