RedPrince
  • RedPrince
What is the integral of ∫(cos x)^2.sqrt{sin x}dx
Mathematics
schrodinger
  • schrodinger
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lgbasallote
  • lgbasallote
\((\cos x)^2 = \cos^2 x = 1 - \sin ^2 x\) i think this can be applied? but seems no :/
RedPrince
  • RedPrince
@lgbasallote It is not working. I already checked it.....!!
anonymous
  • anonymous
|dw:1336216786369:dw|

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anonymous
  • anonymous
is dis a question?
anonymous
  • anonymous
Maybe it is doable with the Weierstrass substitution?
RedPrince
  • RedPrince
@simran yes it is aquestion
RedPrince
  • RedPrince
@estudier what you want to say.... explain please
RedPrince
  • RedPrince
|dw:1336347046770:dw|
anonymous
  • anonymous
Weierstrass sub sin x = 2t/(1+t^2) cos x = (1-t^2)/(1+t^2) dx = 2 dt/(1+t^2)
anonymous
  • anonymous
\[2(7\sin ^{3/2}x-3\sin ^{7/2})/21\]
RedPrince
  • RedPrince
\[\int\limits (cosx)^2\sqrt{sinx }dx= \int\limits (1-t^2)/(1+t^2) 2/(1+t^2) dx\] @estudier this is you want to say!
RedPrince
  • RedPrince
@badi explain please
anonymous
  • anonymous
One tip @RedPrince To write as fractions in latex Use the folowing syntax frac{numerator}{denominator}
RedPrince
  • RedPrince
@shivam_bhalla sorry I am a new user on this site! I will keep it in my mind in future. thanks for tip
anonymous
  • anonymous
No problem. :). trying to solve your problem :D
RedPrince
  • RedPrince
@shivam_bhalla I tried my best but failed!
anonymous
  • anonymous
@RedPrince , the problem is solved. First write \[\cos^2 x = \frac{(1+\cos(2x))}{2}\] We get \[\frac{1}{2}(\int\limits_{}^{}{\sqrt{sinx}dx} +\int\limits_{}^{}{\cos(2x){sinx}dx} )\] Now we integate them separately 1)\[\int\limits_{}^{}{\sqrt{sinx}dx} \] Take \[\sin x = t^2\] Therefore \[cosx dx=2tdt\] \[dx =\frac{ 2t}{cosx}\] Now write \[cosx=\sqrt{1-\sin^2 x}=\sqrt{1-t^4}\] Finally you get \[\int\limits_{}^{} \sqrt{sinx}dx = \int\limits_{}^{}\frac{2tdt}{\sqrt{1-t^4}}\] Now you should be able to integrate this. 2)\[\int\limits_{}^{}(\cos(2x))sinxdx=\int\limits_{}^{}(2\cos^2 x-1)sinxdx\] Now take p=cos x dp= -sinx dx \[-\int\limits_{}^{}{(2t^2-1)}dt\] Now you can integrate this too. Integrate both and finally subtitute t an p in terms of x and simplify. It is pretty long but the only solution which comes to my mind
anonymous
  • anonymous
Oops in part 2 of the problem, it should be \[-\int\limits\limits_{}^{}{(2p^2-1)}dp\]
RedPrince
  • RedPrince
@shivam_bhalla Ok w8 I try it
experimentX
  • experimentX
looks like we need something called elliptic integrals http://www.wolframalpha.com/input/?i=integrate+sqrt%28sinx%29
anonymous
  • anonymous
@experimentX , I am getting the answer with my method :D
experimentX
  • experimentX
\( \int \cos 2x \sqrt{\sin x} dx \) <---- i guess there was error here in your method.
RedPrince
  • RedPrince
@shivam_bhalla I think you did a mistake! there is: \[\cos^2(2x).\sqrt{sinx}\] instead of \[\cos^2(2x).sinx\]
anonymous
  • anonymous
Yes. Sorry for the mistake in part 2 of the problem. let me correct it and tell you
anonymous
  • anonymous
Did anybody try Weierstrass substitution?
RedPrince
  • RedPrince
@shivam_bhalla it's ok!!! I am waiting and try Weierstrass substitution as @estudier is said
RedPrince
  • RedPrince
I try Weierstrass substitution
anonymous
  • anonymous
@estudier , I tried but getting some unsolvable integral. why don't you try and see :D
experimentX
  • experimentX
something doesn't seem right http://www.wolframalpha.com/input/?i=integrate+2x%2Fsqrt%281+-+x^4%29
anonymous
  • anonymous
@experimentX , that is the solution for part 1. Now I am stuck at part 2 of the integration
anonymous
  • anonymous
@experimentX , your initial doubt is right. The solution for 2nd part of intergal is http://www.wolframalpha.com/input/?i=%E2%88%AB%28cos%282x%29%29%28sqrt%28sin%28x%29%29%29dx
RedPrince
  • RedPrince
@shivam_bhalla I already said that I tried my best but failed I black 10 papers for this question! yes @experimentX this is solution for the first part! I continue to solve this question
anonymous
  • anonymous
@experimentX ,Looks like we need to learn elliptical integral.
experimentX
  • experimentX
@shivam_bhalla i think you need to sleep :D :D http://www.wolframalpha.com/input/?i=integrate+2x^2%2Fsqrt%281+-+x^4%29 you missed t from \( \sqrt{\sin x}\) in the first part
anonymous
  • anonymous
LOL. looks like I really need to sleep now. Sorry @RedPrince .@experimentX .
RedPrince
  • RedPrince
@shivam_bhalla it's ok dear text my friend and he reply that he solve it! when I get answer I will post it!
experimentX
  • experimentX
sure ... let me know too.
anonymous
  • anonymous
@estudier ,Wolfram is getting this by Weierstrass substitution http://www.wolframalpha.com/input/?i=%E2%88%AB%282sqrt%282%29%29%281-t^2%29^2sqrt%28t%29%2F%281%2Bt^2%29^%285%2F2%29++dt @experimentX , check it please
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=integrate+sqrt%28x++-+x^3%29+dx
experimentX
  • experimentX
is it indefinite integral??
anonymous
  • anonymous
@shivam_bhalla Yes,seems like elliptic integral is the way...
anonymous
  • anonymous
Although I can't help thinking there ought to be a substitution that works Where is this question from?
RedPrince
  • RedPrince
this is my assignment and I post it!
anonymous
  • anonymous
@RedPrince , I am sure there must be some typo in the question. BTW, in which grade are you studying in? (or What's your age?)
RedPrince
  • RedPrince
I am a student of BS Computer Science and I am 19 years old!
anonymous
  • anonymous
Ohh. LOL I am just 17 and just completed my 12th grade
anonymous
  • anonymous
Maybe it needs two substitutions. I am surprised that they gave you such an integral.....
RedPrince
  • RedPrince
@estudier why you are surprised! I am a university student not a college student
anonymous
  • anonymous
@estudier , I doubt so. I have tried various types of substitutions (till my level best) Until today, I never heard of elliptic integral
RedPrince
  • RedPrince
@shivam_bhalla me also first time I heard about elliptic integral
anonymous
  • anonymous
@satellite73 Anything spring to mind?
anonymous
  • anonymous
lol university, not college. is the university not made up of colleges?
anonymous
  • anonymous
LOL @satellite73
anonymous
  • anonymous
no i am braid dead and i loathe techniques of integration. that is why they invented computers
anonymous
  • anonymous
lol
anonymous
  • anonymous
Ok. @amistre64 should help when he comes online :)
anonymous
  • anonymous
@amistre64 's profile picture itself is integration :P
anonymous
  • anonymous
(u^(1/2)(1-u^2)) du/(1-u^2)^(1/2) Best I can do, still gives elliptic
anonymous
  • anonymous
@RedPrince , did you reconfirm the question ??
RedPrince
  • RedPrince
@shivam_bhalla Yes I asked to my teacher and he reconfirmed the question as: \[\int\limits \cos^3x \sqrt sinx dx\]
experimentX
  • experimentX
is there limits?? or just indefinite integral??
experimentX
  • experimentX
looks like this is quite easier that the original question http://www.wolframalpha.com/input/?i=integrate+cos^3x+sqrt%28sinx%29
anonymous
  • anonymous
@RedPrince Yes, same question as @experimentX
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=integrate+%281-x^2%29sqrt%28x%29 let sin(x) = u, cos^2x = 1 - u^2, cos(x)dx = du
anonymous
  • anonymous
I think it would be better if t^2 = sinx 2tdt = cos x dx \[\int\limits_{}^{}(1-t^4)(t)(2t )(dt)\]
experimentX
  • experimentX
well ... that solves our problem!!
anonymous
  • anonymous
Good work @experimentX :) By the way, sometype wrong question teaches us new things :P :D
anonymous
  • anonymous
*sometimes
experimentX
  • experimentX
yea elliptical integrals!!
RedPrince
  • RedPrince
@shivam_bhalla yes sometimes they teaches us many things that are helpful in future....!! well thanks a lot all of you guys for helping me!!! espacially @experimentX ,@estudier and @shivam_bhalla
anonymous
  • anonymous
\[\frac{2}{5} \left(\sin ^{\frac{3}{2}}(x) \cos (x)-2 E\left(\left.\frac{1}{4} (\pi -2 x)\right|2\right)\right) \] This the answer I got from Mathematica. It does not seem to have an easy closed form
RedPrince
  • RedPrince
@eliassaab how this answer is come
anonymous
  • anonymous
This integral does not have a closed form in terms of usual functions, otherwise, mathematica would have found it.
anonymous
  • anonymous
|dw:1336919117275:dw|
anonymous
  • anonymous
@mahmit2012 , yes. @RedPrince said there was typo in original question he first mentioned
anonymous
  • anonymous
|dw:1336919188247:dw|
RedPrince
  • RedPrince
@mahmit2012 yes you are true but how you get it!

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