Integral Calculus Question 204
Question: What is $ \int{{{\tan }^{2}}x{{\sec }^{4}}x,dx} $ equal to?
Options:
A) $ \frac{{{\sec }^{5}}x}{5}+\frac{{{\sec }^{3}}x}{3}+c $
B) $ \frac{{{\tan }^{5}}x}{5}+\frac{{{\tan }^{3}}x}{3}+c $
C) $ \frac{{{\tan }^{5}}x}{5}+\frac{{{\sec }^{3}}x}{3}+c $
D) $ \frac{{{\sec }^{5}}x}{5}+\frac{{{\tan }^{3}}x}{3}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let $ I=\int{{{\tan }^{2}}x{{\sec }^{4}}xdx} $ Let $ \tan x=t $
$ \Rightarrow {{\sec }^{2}}xdx=dt $
$ \therefore I=\int{{{\tan }^{2}}x.{{\sec }^{2}}x.{{\sec }^{2}}x.dx} $ $ =\int{{{\tan }^{2}}x(1+{{\tan }^{2}}x){{\sec }^{2}}x.dx} $
$ \therefore I=\int{t^{2}(1+t^{2})dt=\int{(t^{2}+t^{4})dt}} $ $ =\frac{t^{5}}{5}+\frac{t^{3}}{3}+c=\frac{{{\tan }^{5}}x}{5}+\frac{{{\tan }^{3}}x}{3}+c $
BETA
