Q. If $I_{n} = \int_{}^{}{\tan^{n}xdx}$ for $n \geq 2$, then $I_{n} + I_{n - 2}$ is equal to
A) $\tan^{n}{x + c}$
B) $\frac{(\tan^{n - 1}x)}{n - 1} + c$
C) $\frac{\tan^{n}x}{n} + c$
D) $n\tan^{n}x + c$
Explanation
(b)
$I_{n} = \int_{}^{}{\tan^{n}xdx} = \int_{}^{}{\tan^{n - 2}x(\sec^{2}x - 1)dx}$
$= \int_{}^{}{\tan^{n - 2}x.\sec^{2}xdx - I_{n - 2}}$
$\Rightarrow \ I_{n} + I_{n - 2} = \int_{}^{}{\tan^{n - 1}x.\sec^{2}x}dx$
Put $\tan{x = t\ \ \ \Rightarrow \ \ \sec^{2}xdx = dt}$
$\therefore\ I_{n} + I_{n - 2} = \int_{}^{}t^{n - 2}dt = \frac{t^{n - 1}}{n - 1} + c$
$\Rightarrow \ I_{n} + I_{n - 2} = \frac{(\tan^{n - 1}x)}{n - 1} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.