Solve integral by trig substitution

Somniantis

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Feb 11, 2010
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Hello board members,
I need help with an integral. The chapter is from is "Solving integrals by trigonometric substitutions", so I tried that, but got stuck.

?{[(1-X[sup:2etpegdz]2[/sup:2etpegdz])[sup:2etpegdz]5/8[/sup:2etpegdz]]/X[sup:2etpegdz]8[/sup:2etpegdz]}dx

I tried
x=siny, dx=cosy dy so the integral would become:
?{[(1-sin[sup:2etpegdz]2[/sup:2etpegdz]y)[sup:2etpegdz]5/8[/sup:2etpegdz]]cosy/sin[sup:2etpegdz]8[/sup:2etpegdz]y}dy= ?[cos[sup:2etpegdz]9/4[/sup:2etpegdz]y/sin[sup:2etpegdz]8[/sup:2etpegdz]y] dy, but got stuck from there.
I also tried the substitution x= tanh y, and work with tanh and sech instead of cos and sin, still no luck.

Any help appreciated. Thanks.
 
Somniantis said:
Hello board members,
I need help with an integral. The chapter is from is "Solving integrals by trigonometric substitutions", so I tried that, but got stuck.

?{[(1-X[sup:lsp6jjg2]2[/sup:lsp6jjg2])[sup:lsp6jjg2]5/8[/sup:lsp6jjg2]]/X[sup:lsp6jjg2]8[/sup:lsp6jjg2]}dx

I tried
x=siny, dx=cosy dy so the integral would become:
?{[(1-sin[sup:lsp6jjg2]2[/sup:lsp6jjg2]y)[sup:lsp6jjg2]5/8[/sup:lsp6jjg2]]cosy/sin[sup:lsp6jjg2]8[/sup:lsp6jjg2]y}dy= ?[cos[sup:lsp6jjg2]9/4[/sup:lsp6jjg2]y/sin[sup:lsp6jjg2]8[/sup:lsp6jjg2]y] dy, but got stuck from there.
I also tried the substitution x= tanh y, and work with tanh and sech instead of cos and sin, still no luck.

Any help appreciated. Thanks.

If you posted the problem correctly - somebody does not love you....

Running it through Wolfram-alpha, I get,

integral {(1-X^2)^(5/8)/X^8} dX = {(35 X^8[sub:lsp6jjg2]2[/sub:lsp6jjg2]F[sub:lsp6jjg2]1[/sub:lsp6jjg2](3/8,1/2;3/2;X^2)+4 (1-X^2)^(5/8) (35 X^6+20 X^4+16 X^2-64))/(1792 X^7)}+constant

where [sub:lsp6jjg2]2[/sub:lsp6jjg2]F[sub:lsp6jjg2]1[/sub:lsp6jjg2] is hypergeometric function (and I do not know how to interpret that)
 
\(\displaystyle \int \frac{(1-x^{2})^{\frac{5}{8}}}{x^{8}}dx\)

You're correct, SK. This integral is not elementary. Are you sure it is typed correctly?. I ran it through Mathematica and got something involving Hypergeometric. Therefore, it is beyond the scope of a normal trig sub course.
 
Okay, now that you guys said something was wrong with it, I called my instructor and double checked...
(runs and hides in shame...!!!)...

Sorry guys,it seems I had copied a wrong power...I feel like a brainless turd(especially after I fought with the wrong problem for a couple of hours...)
...it should be ?{[(1-X[sup:1v5cmtxm]2[/sup:1v5cmtxm])[sup:1v5cmtxm]5/2[/sup:1v5cmtxm]]/X[sup:1v5cmtxm]8[/sup:1v5cmtxm]}dx, which results in :

?{(cos[sup:1v5cmtxm]2[/sup:1v5cmtxm]y)[sup:1v5cmtxm]5/2[/sup:1v5cmtxm]cos y/sin[sup:1v5cmtxm]8[/sup:1v5cmtxm]y}dy = ? cot[sup:1v5cmtxm]6[/sup:1v5cmtxm]y csc[sup:1v5cmtxm]2[/sup:1v5cmtxm]y dy, which should be easy, -1/7 cot[sup:1v5cmtxm]7[/sup:1v5cmtxm]y

Thanks anyone for trying and sorry about that!!
A good way to start in a new forum,eh?~))
 
Somniantis said:
Okay, now that you guys said something was wrong with it, I called my instructor and double checked...
(runs and hides in shame...!!!)...

Sorry guys,it seems I had copied a wrong power...I feel like a brainless turd(especially after I fought with the wrong problem for a couple of hours...)
...it should be ?{[(1-X[sup:3s8t72fl]2[/sup:3s8t72fl])[sup:3s8t72fl]5/2[/sup:3s8t72fl]]/X[sup:3s8t72fl]8[/sup:3s8t72fl]}dx, which results in :

?{(cos[sup:3s8t72fl]2[/sup:3s8t72fl]y)[sup:3s8t72fl]5/2[/sup:3s8t72fl]cos y/sin[sup:3s8t72fl]8[/sup:3s8t72fl]y}dy = ? cot[sup:3s8t72fl]6[/sup:3s8t72fl]y csc[sup:3s8t72fl]2[/sup:3s8t72fl]y dy, which should be easy, -1/7 cot[sup:3s8t72fl]7[/sup:3s8t72fl]y

Thanks anyone for trying and sorry about that!!
A good way to start in a new forum,eh?~))

don't forget to "back-substitute" and convert the answer to "X". Also don't forget the integration constant...
 
\(\displaystyle Ok, \ f(x) \ = \ \int \frac{(1-x^{2})^{5/2}}{x^{8}}dx \ = \ \bigg(\frac{-1}{7}\bigg)\bigg[\frac{(1-x^{2})^{7/2}}{x^{7}}\bigg] \ + \ C\)

\(\displaystyle Now, \ what \ is \ the \ domain \ of \ f(x)?\)
 
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