Mathematics

$$\displaystyle \int x\sec x^{2}dx$$ is equal to


ANSWER

$$\displaystyle \frac{1}{2}\log \left ( \sec x^{2}+\tan x^{2} \right )+k$$


SOLUTION
$$\displaystyle \int x\sec x^{2}dx$$

Put $$x^2=t$$
$$\Rightarrow 2xdx=dt$$

So,
$$\displaystyle I=\frac{1}{2}\int  \sec  tdt$$

$$\displaystyle I=\frac{1}{2} \log|\sec t+\tan t|+k$$

$$\displaystyle I=\frac { 1 }{ 2 } \log  |\sec { x^{ 2 } } +\tan  x^{ 2 }|+k$$
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Single Correct Medium Published on 17th 09, 2020
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