## RedPrince 4 years ago what is the result of this question: sqrt{1-t^4}.2tdt

1. RedPrince

$\int\limits \sqrt{1-t^4}.2tdt$

2. .Sam.

$u=t^2~~~and ~~~d u=2 td t$

3. anonymous

Use substitution: Let t^2 = m then 2tdt = dm The new integral is; $\int\sqrt{1-m^2}dm$

4. .Sam.

$\int\limits \sqrt{1-u^2} \, du$ u=sin(a) $\sqrt{1-u^2}\text{ = }\sqrt{1-\sin ^2(s)}\text{ = }\cos (s)\text{ and }s=\sin ^{-1}(u)$

5. .Sam.

$\int\limits \cos ^2(s) \, ds$ half angle $\int\limits \left(\frac{1}{2} \cos (2 s)+\frac{1}{2}\right) \, ds$ separate $\int\limits \frac{1}{2} \, ds+\frac{1}{2}\int\limits \cos (2 s) \, ds$ solve

6. RedPrince

you are right but I obtained it from: $\int\limits \cos^2x \sqrt{sinx} dx$

7. .Sam.

?

8. RedPrince

@.Sam. don't understand how I get it!

9. RedPrince

@shivam_bhalla help me dear!

10. .Sam.

i didn't get $\int\limits\limits \cos^2x \sqrt{sinx} dx$

11. anonymous

@RedPrince , follow what @.Sam. told

12. RedPrince

@.Sam. integrate it! I get the result $\int\limits \sqrt{1-t^4}.2tdt$ from $∫\cos^2x \sqrt{sinx}dx$ by substituting $sinx=t^2$

13. RedPrince

@shivam_bhalla I also follow. you also?

14. .Sam.

its a long step to integrate this $∫\cos^2x \sqrt{sinx}dx$

15. anonymous

I simply remember that $\int\limits_{}^{}\sqrt{a^2-x^2} = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}{(x/a)}$

16. RedPrince

@.Sam. you guide me from where I ask question! see my uestion

17. anonymous

@RedPrince , see the above integral I wrote, substitute and simplify

18. RedPrince

@shivam_bhalla this is good and now the question is solved.....!! woooowwwwWWw. Thanksss!

19. .Sam.

you could use this too

20. anonymous

@.Sam. , was just writing the same :)

21. .Sam.

or

22. RedPrince

@.sam yes you are right. this is also helpful.

23. .Sam.

but that's just the formula without steps

24. anonymous

@RedPrince , I suggest you remember these formulas because they take a long time to derive. This is just formulas to quickly solve problems

25. anonymous

But , however I suggest you try to derive these formulas on your own first and then remember them

26. RedPrince

@.Sam. it doesn't matter. I will prove them myself. again thanks!

27. RedPrince

@shivam_bhalla yes, I will remember these formulas Inshallah.

28. asnaseer

RedPrince: I /think/ you may have made a mistake in your substitutions. I get this after the substitutions:$\int\limits \sqrt{1-t^4}.2t^2dt$

29. asnaseer

$\sin(x)=t^2$therefore:$\cos(x)dx=2tdt$

30. asnaseer

$\cos^2(x)=1-\sin^2(x)=1-t^4$

31. asnaseer

therefore:$\cos^2(x)\sqrt{sin(x)}dx=\cos^2(x)t*\frac{2tdt}{cos(x)}=\cos(x)2t^2dt$$\qquad=\sqrt{1-t^4}2t^2dt$

32. anonymous

@asnaseer , hawk eye :)

33. asnaseer

:)

34. anonymous

So our new integral should be $m=t^2$ $dm = 2tdt$ $\int\limits_{}^{}\sqrt{1-m^2}\sqrt{m}dm$

35. asnaseer

I /believe/ the original integral is not as straightforward as it might look. this is what wolfram thinks the answer should be: http://www.wolframalpha.com/input/?i=integrate+cos^2%28x%29*sqrt%28sin%28x%29%29

36. RedPrince

@asnaseer yes, you are right but how can I solve it i.e., $\int\limits \sqrt{m-m^3}$

37. asnaseer

see the link I posted - it is not an easy integral (even in this form): http://www.wolframalpha.com/input/?i=integrate+sqrt%28x-x^3%29

38. asnaseer

you need an expert like @JamesJ - he might be able to see a way to getting to the answer.

39. RedPrince

@asnaseer O.K., just see who help me in solving this question!

40. anonymous

@RedPrince , the qustion is mis understood. the first solution is the correct one

41. asnaseer

@vamgadu have you read ALL the comments above? the original equation to solve was:$∫\cos^2x \sqrt{sinx}dx$this was incorrectly transformed to:$\int\limits \sqrt{1-t^4}.2tdt$

42. RedPrince

43. RedPrince

@heena @amistre64 @AccessDenied @satellite73 help plaese if you have idea!

44. amistre64

just integrate by parts

45. amistre64

(1+t^4)^1/2 2t ..... hmm, seems i confused my integrating and deriving terms again -2

46. amistre64

its still by parts but a trig sum does seem plausible

47. amistre64

let tan(u) = t^2 sec^2(u) du = 2t dt replace the substitutions$\int \sqrt{1+tan^2(u)}\ du$

48. amistre64

really need to wake up this morning :) $\int \sqrt{1+tan^2(u)}\ sec^2(u)du$

49. amistre64

that would have worked out if I hadnt misread the operator; but same concept using the a sin or cos instead of a tangent