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
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