# Integral of a fuction

#### amitzr3

##### New member
Hey, could'nt find a solution for this.

I also attached my try.

Help would be appreciated.

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#### Subhotosh Khan

##### Super Moderator
Staff member
Apply:
$$\displaystyle u = sin^{-1}(x)$$

$$\displaystyle \frac{d}{dx}[sin^{-1}(x)] \ = \ \dfrac{1}{\sqrt{1-x^2}}$$

so your integral becomes (1/u) du.....

That should help

#### HallsofIvy

##### Elite Member
That looks much too complicated! Whoever gave you this problem clearly expects that you have previously learned that the derivative of $$\displaystyle y= \arcsin(x)$$ is $$\displaystyle y'= \frac{1}{\sqrt{1- x^2}}$$ so an obvious substitution is $$\displaystyle u= \arcsin(x)$$. Then $$\displaystyle du= \frac{dx}{\sqrt{1- x^2}}$$ and the integral becomes $$\displaystyle \int\frac{du}{u}$$ which should be easy.