## anonymous 4 years ago in Substitution Rule

1. anonymous

2. myininaya

$\int\limits\limits_{0}^{\frac{1}{2}}\frac{\sin^{-1}(x)}{\sqrt{1-x}} dx$

3. anonymous

i would make the substitution: x = sin^2(u) dx = 2sin*cos du $\rightarrow 2\int\limits_{0}^{.5}u^{2} \sin(u) du$ Then use integration by parts

4. anonymous

but I don't know how to do it

5. myininaya

Do you know integration by parts?

6. anonymous

yeah i do but that question is on substitution not in integration by parts

7. myininaya

Why can't we do both?

8. anonymous

you can do :)

9. anonymous

it

10. myininaya

$\int\limits_{}^{}u^2 \sin(u) du=u^2(- \cos(u))-\int\limits_{}^{}2u (-\cos(u)) du$ you will need to do integration by parts one more time

11. anonymous

ok Thank you a lot guys :)

12. TuringTest

one thing I am confused on here, where did we get u^2 ? are you saying arcsin(sin^2u)=u^2 ?

13. anonymous

yes arcsin(sin u) = u by definition of the inverse function

14. TuringTest

somehow the idea that that extends to arcsin(sin^n(u))=u^n has never been brought to my attention. thanks

15. anonymous

could we have just used the substitution:$u = \sin^{-1}(x)$

16. anonymous

ah disregard, i didnt look at the denominator properly.

17. myininaya

yeah i don't know if that is true turing

18. TuringTest

yeah it can't be

19. TuringTest

arcsin(sin(u^n))=u^n arcsin(sin^n(u))=?

20. myininaya

i agree with the first equation you have turing

21. anonymous

wait now that i look at it again, i think i have it backwards. sorry everyone sin^2(arcsin u) = u^2 , thats what i thought it was

22. anonymous

I got

23. TuringTest

Oh man, I thought I was loosing it :/

24. anonymous

i got |dw:1327181038255:dw|

25. anonymous

26. TuringTest

wolfram disagrees http://www.wolframalpha.com/input/?i=integral+from+0+to+0.5+%28arcsin%28x%29%2Fsqrt%281-x%29 but I don't trust wolfram anymore, lol

27. anonymous

I don't know man that question is hard to me :(

28. myininaya

is it written correctly on the file you posted?

29. TuringTest

yeah, if there was an x^2 in the denom our lives would be about a million times easier

30. anonymous

let me check it again, I may write it wrong

31. anonymous

yeah you were right TuringTest I'm really sorry guys

32. anonymous

i'm sorry guys there is X^2

33. TuringTest

$\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$so $u=\sin^{-1}x\to du=\frac{1}{\sqrt{1-x^2}}dx$giving$\int udu$

34. anonymous

yeah that what i did so far , but i wanna know if i'm doing right TuringTest. Can you do until the end of it ?

35. TuringTest

$\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$I'm going to be lazy and not use the proper bound notation, just plug numbers in at the end. $u=\sin^{-1}x\to du=\frac{1}{\sqrt{1-x^2}}$giving$\int udu=\frac{u^2}{2}=\frac{1}{2}(\sin^{-1}x)^2|_{0}^{1/2}=\frac{1}{2}[(\sin^{-1}(1/2))^2-(\sin^{-1}(0))^2]$$=\frac{1}{2}[(\frac{\pi}{6})^2-0]=\frac{\pi^2}{72}$

36. anonymous

Thank You so much Turing that was very helpful

37. TuringTest

anytime :)