### JEE Main Solved Paper 2017 Question 17

##### Question: Let $ I _{n}=\int _{{}}^{{}}{{{\tan }^{n}}xdx,(n>1).I _4+I _6} $ $ =a{{\tan }^{5}}x+bx^{5}+C, $ where C is a constant of integration, then ordered pair (a, b) is equal to:- [JEE Main Solved Paper-2017] ]

#### Options:

A) $ ( -\frac{1}{5},0 ) $

B) $ ( -\frac{1}{5},1 ) $

C) $ ( \frac{1}{5},0 ) $

D) $ ( \frac{1}{5},-1 ) $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

$ I _4+I _6=\int _{{}}^{{}}{({{\tan }^{4}}x+{{\tan }^{6}}x)},dx=\int _{{}}^{{}}{{{\tan }^{4}}x{{\sec }^{2}}x,dx} $

$ =\frac{1}{5}{{\tan }^{5}}x+c\Rightarrow a=\frac{1}{5},b=0 $