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