C4 Integration Question - Help! Watch

Kittyyyyyyy
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#1
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I'm not too good with integration, but can anyone help me with integrating this?

\displaystyle\int^\frac{1}{2}_0\
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silent ninja
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I cant see anything?
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musicmobile
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Whats the bit after that? the bit your meant to intergrate!?
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Kittyyyyyyy
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sorry - clicked post instead of preview :P
will post the revised version in a minute
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Andrew28913
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Never have I seen quite so thrilling an integral.
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Kittyyyyyyy
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Here it is -

\displaystyle\int^\frac{1}{2}_0\ \frac{1}{(1-x^2)^\frac{3}{2}}\ dx
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Notnek
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Try a substitution x=sin(u).
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silent ninja
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you need to use a substitution, x=sinu or cosu will work
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Kittyyyyyyy
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I don't understand how to do it though?
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Notnek
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(Original post by Kittyyyyyyy)
I don't understand how to do it though?
Where have you got up to. Post any ideas/working you have.
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Creative
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dont you just integrate by recognition

to get x/sqrt(1-x^2)
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Notnek
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(Original post by Creative)
dont you just integrate by recognition

to get x/sqrt(1-x^2)
No - that's not right.
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bruceleej
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(Original post by Kittyyyyyyy)
I don't understand how to do it though?
what you do is, replace the x in the integrand with sinu and then for dx. differentiate x = sin u with respect to u, and then make dx the subject and replace dx with whatever your result was.
For the limits, find u when x= 0 and at x=1/2. Then simplify things and the integral should be simple to evaluate from there.
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silent ninja
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(Original post by Creative)
dont you just integrate by recognition

to get x/sqrt(1-x^2)
Why don't you differentiate to check your answer?
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Kittyyyyyyy
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I don't know how to start it or anything, being incredibly thick at integration.
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M.A.H
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Start off using the substitution

x=sinu

and then differentiate that and see what you can replace
so you have

dx= cosu.du
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Creative
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(Original post by silent ninja)
Why don't you differentiate to check your answer?
yeah i get the original integral if I do that

( unless i'm doing something wrong :P ), used product rule to differentiate.
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Notnek
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I'll start you off:

If you substitute x=sinu you get:

\displaystyle \int^{x=\frac{1}{2}}_{x=0} \frac{1}{(1-sin^2(u))^{\frac{3}{2}}} dx = \int^{x=\frac{1}{2}}_{x=0} \frac{1}{(cos^2(u))^{\frac{3}{2}  }} dx = \int^{x=\frac{1}{2}}_{x=0} \frac{1}{cos^3(u)} dx

Now find \frac{dx}{du} to find dx in terms of du and substitute...
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silent ninja
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(Original post by Creative)
yeah i get the original integral if I do that

( unless i'm doing something wrong :P ), used product rule to differentiate.

Product or quotient rule, you are doing something wrong :p:

Op, have you done any integration by substitution before?
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Kittyyyyyyy
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#20
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Using x=sin\theta could I just do the same?
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