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On solving the equation in the photo by 2 methods I'm getting two answers. one with a complex number and the other without it. Can someone explain me why??[enter image description here

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    Writing equalities involving indefinite integrals leads possibly to mistakes like this one. You should realize that the functions you are integrating are to be defined on some interval and also to be of class $C^1$ if you want to integrate by parts. Here you can choose to work on $(-1,0)$ or on $(0,1)$ because of the presence of $arcsin(x)$ and because of division by $x^2$. So switching to $\sqrt{x^2-1}$ is illicit ! – 2017-02-12
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    y is switching to (x^2 - 1)^1/2 illicit? @Adren – 2017-02-12
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    Method two with some changes likes good. – 2017-02-12
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    @DhruvRaghunath: yes because both expressions $\sqrt{1-x^2}$ and $\sqrt{x^2-1}$ are not defined for the same values of $x$ ... In fact, you are not computing twice the same thing, but first for $x$ in some interval and then for $x$ in another interval, disconnected from the previous one. – 2017-02-12
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    @DhruvRaghunath: As I wrote in a previous comment, the first expression you get is valid for $x\in I$ where the interval $I$ is such that $I\subset(-1,0)\cup(0,1)$ and the second one for $x\in I$ where $I\subset(-\infty,-1)\cup(1,+\infty)$. – 2017-02-12
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    BTW, => doesn't mean =. – 2017-02-12

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