Integral Calculus Question 389
Question: $ \int_{{}}^{{}}{\frac{1}{\sqrt{x}}{{\tan }^{4}}\sqrt{x}}{{\sec }^{2}}\sqrt{x}\ dx= $
Options:
A) $ 2{{\tan }^{5}}\sqrt{x}+c $
B) $ \frac{1}{5}{{\tan }^{5}}\sqrt{x}+c $
C) $ \frac{2}{5}{{\tan }^{5}}\sqrt{x}+c $
D) None of these
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Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{1}{\sqrt{x}}{{\tan }^{4}}\sqrt{x},.,{{\sec }^{2}}\sqrt{x},dx} $ Put $ \tan \sqrt{x}=t\Rightarrow \frac{{{\sec }^{2}}\sqrt{x}}{2\sqrt{x}},dx=dt, $ then it reduces to $ 2\int_{{}}^{{}}{t^{4}dt}=\frac{2}{5}{{(\tan \sqrt{x})}^{5}}+c=\frac{2}{5}{{\tan }^{5}}\sqrt{x}+c $ .
BETA
