Mathematics

# $\displaystyle\int {\dfrac{{{{\tan }^4}\sqrt x \,{{\sec }^2}\sqrt x }}{{\sqrt x }}} \,dx$

##### SOLUTION
$I=\displaystyle \int \dfrac {\tan^4\sqrt x.\sec^2\sqrt {x}}{\sqrt x}dx$
Let $\tan \sqrt x=t$
$\sec^2\sqrt {x}.\dfrac {1}{2\sqrt x}dx =dt$
$\Rightarrow \ \dfrac {\sec^2\sqrt x}{\sqrt x}dx=edt$
$\therefore \ I=\displaystyle \int t^{4}.2dt =2\displaystyle \int t^4dt$
$=\dfrac {2}{5}t^5$, substituting $t=\tan \sqrt x$
We get $I=\dfrac {2}{5}\tan^5 \sqrt x+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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