Mathematics

The value of integratin $$\displaystyle \int x\sec^{-1}xdx$$ is


ANSWER

$$\displaystyle \frac{1}{2}x^{2}\sec^{-1}x-\frac{1}{2}\sqrt{x^{2}-1}+c$$


SOLUTION
The value of $$\displaystyle \int x  \sec^{-1}x  dx$$ is
$$\displaystyle =\sec^{-1}x\cdot \dfrac {x^2}{2}-\displaystyle \int \dfrac {1}{\sqrt {x^2-1}}\cdot \dfrac {x^2}{2}dx$$
$$\displaystyle =\dfrac {1}{2}x^2\sec^{-1}x-\dfrac {1}{4}\displaystyle \int \dfrac {dx^2}{\sqrt {x^2-1}}dx$$
$$\displaystyle =\dfrac {1}{2}x^2\sec^{-1}x-\dfrac {1}{2}\sqrt {x^2-1}+c$$
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Single Correct Medium Published on 17th 09, 2020
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