Trignometric substitutio of the indefinite integral x^3*sqrt(16+x^2)dx

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Trignometric substitutio of the indefinite integral x^3*sqrt(16+x^2)dx

Mathematics
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just a sec k?
k

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so far this is what i have
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kk looks good so far
http://answers.yahoo.com/question/index?qid=20080508091701AAA0Mq0 im fixing to have dinner this link should help you with the rest im sorry
the answer should be [(16-x^2)^5/2]/5-[16(16+x^2)^3/2]/3+C
you should also put your answer back in terms of x i might be back on later if you haven't figured it out by then k?
k thanks for your help.
im sorry hai peace
can you please help me?
amistre its really tough
x^3*sqrt(16+x^2) dx; yeah, it can be..
4tan(t) = x^2
ack!!.. 4tan(t) = x
64 tan^3(t) * 4 sec(t) dt ; just gotta determine the dt now
its 4sec^2 theta dtheta
4sec^2(t) dt = dx right?
yea
64 tan^3(t) * 16sec^3(t) dt then right?
yea
i put the answer above for reference, this problem is driving me crzy
1024 (tan(t)sec(t))^3
can you do it on your own and see if you get that?
http://www.wolframalpha.com/input/?i=int%2864+tan^3%28t%29+*+16sec^3%28t%29%29+dt
I aint got the time tonight, libraries closing soon but wolfram gives steps
http://www.wolframalpha.com/input/?i=int%28x^3+sqrt%2816%2Bx^2%29%29dx or this for the original
You've got it. You've worked your way to a function that can be integrated.
Integral (sec theta tan theta)^3=sec^3 theta. Now you have to convert back. Good job Amistre. He just had to run and didn't have time to interpret it.

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