Mathematics

$$\displaystyle \int \dfrac{1}{\sqrt{x}} \tan^4 \, \sqrt{x} \, \sec^2 \, \sqrt{x} \, dx = $$


ANSWER

$$\dfrac{2}{5} \tan^5 \, \sqrt{x} + c$$


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

substitute $$u=\tan \sqrt{x}\rightarrow du=\dfrac{\sec ^2\sqrt{x}}{2\sqrt{x}}dx$$

$$=2\int u^4du$$

$$=\dfrac{2}{5}u^5$$

$$=\dfrac{2}{5}\tan ^5 \sqrt{x}+C$$
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Single Correct Medium Published on 17th 09, 2020
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