Integral Calculus Question 441
Question: $ \int_{{}}^{{}}{\tan x}{{\sec }^{2}}x\sqrt{1-{{\tan }^{2}}x}\ dx= $
Options:
A) $ -\frac{1}{3}{{(1-{{\tan }^{2}}x)}^{3/2}}+c $
B) $ \frac{1}{3}{{(1-{{\tan }^{2}}x)}^{3/2}}+c $
C) $ -\frac{2}{3}{{(1-{{\tan }^{2}}x)}^{2/3}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\tan x,.,{{\sec }^{2}}x\sqrt{1-{{\tan }^{2}}x},dx} $ Put $ \tan x=t\Rightarrow {{\sec }^{2}}x,dx=dt, $ then it reduces to $ \int_{{}}^{{}}{t\sqrt{1-t^{2}},dt} $ Now again, put $ 1-t^{2}=u, $ then its reduced form is $ -\frac{1}{2}\int_{{}}^{{}}{\sqrt{u},du}=-\frac{1}{3}{u^{3/2}}+c=-\frac{1}{3}{{(1-{{\tan }^{2}}x)}^{3/2}}+c. $
BETA
