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Old 04-20-09, 07:41 PM   #1
phantomcow2
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u substitution problem -- integration

Find the indefinite integral for sec^3x*tanx using u substitution.

I've done pretty well solving u substitution problems up to this point, but this problem continues to stump me; there is something I am missing.

let u=tanx
du = sec^2x*dx

I figure that maybe I can "break apart" the sec^3x term into sec^2x*secx, giving me the following:
sec^2x*secx*u*dx

But du=sec^2dx so substituting gives secx*u*du

But I can't find the anti derivative of this
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Old 04-20-09, 11:06 PM   #2
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There was a time when I knew this stuff. Sadly I've forgotten it all. My mind is tuned with boolean algebra and logic circuits ATM. Sorry...
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Old 04-21-09, 06:44 AM   #3
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assuming secx*u*du is correct, the integral to u would be

0.5*secx*u^2+C

To see why, just reverse it. Again, not really a math wiz in the formal sense, so please check yourself. Also, this has been ages.
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Old 04-21-09, 07:18 AM   #4
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Maybe a better way is to write:

sqrt(1+tan(x)^2)^3*tan(x) dx

Because then if u=tan(x) you get:

sqrt(1+u^2)^3*u du --> ((1+u^2)^0.5)^3*u du --> (1+u^2)^1.5*u du

which eliminates the extra x, which I am not sure what to do about (integrate the result to x aftewards?).

Finding the anitderivative should be easy from there on, but is has been ages, so you best do that yourself
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Old 04-21-09, 07:38 AM   #5
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Just a thought:

(1+u^2)^1.5*u du

looks like

h(g(u))'*g(u)'

with g(u)=1+u^2

thus h()'=()^1.5 making H()=(()^2.5)/2.5

((1+u^2)^2.5)/2.5

However, the derivative of 1+u^2 is 2u so apparently there was a separate multiplicative term which ended up as 1/u. Then try to compensate for this.
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