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 $
