## anonymous 3 years ago Evaluate the Intergal

1. anonymous

$\int\limits\limits_{}^{}\frac{ \arcsin(x) }{ x^2 } dx$

2. anonymous

3. anonymous

Not sure either, but what does $$x=\sin(u)$$ get you?

4. anonymous

Why would you do that?

5. AccessDenied

I was thinking about trying integration by parts... have you tried that yet? :)

6. anonymous

It would get rid of the arcsin.

7. anonymous

It's just a guess

8. anonymous

Well I was going to make a substitution and then use intergration by parts but I am not sure...

9. anonymous

What if I said u=arcsin(x) ?

10. anonymous

That's the same as x = sin(u)

11. anonymous

Right. But if I said that I would get a squart root and I could use trigonometric substitution.

12. anonymous

Allright so I so far got... Use intragtion by parts: u=arcsinx) $du=\frac{ 1 }{ \sqrt{1-x^2} }$ dv=x^-2 dx v=-x^-1

13. anonymous

$\frac{ -\arcsin(x) }{ x }-\int\limits_{}^{}\frac{ 1 }{ x \sqrt{1-x^2} }dx$

14. anonymous

I am kinda stuck at this point.

First do the substitution $$x=\sin(u)\implies dx=\cos(u)du$$ then $$\int \frac{\arcsin{x}}{x^2}\;dx=\int \frac{u\cos(u)}{\sin^2(u)}\;du=\int u\csc(u)\cot(u)\;du$$ $$\int u\frac{d(-\csc(u))}{du}du$$ do the integration by parts

16. anonymous

Hmm let me see...

17. anonymous

@BAdhi So I got to this point $\int\limits_{}^{}ucsc(u)\cot(u) du$

18. anonymous

I would use integration by parts here right?

we know that $$\frac{d\csc(u)}{du}=-\csc(u)\cot(u)$$ then, \begin{align*}\int u\csc(u)\cot(u)\,du&=\int u\frac{-d\csc(u)}{du}\,du\\ &=-u\csc(u)+\int \csc(u)\,du \end{align*}

20. anonymous

Yep I got that far.

21. anonymous

Now I need to figure out the integral of csc(x).

22. anonymous

Thanks I got this :) .

23. anonymous

could use x = sin(u)

24. anonymous

Thank you too wio :) .