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I was doing some integrations and one of the problems is: x39x2dx

The answer says: x23(9x2)3/2215(9x2)5/2+C

Here is my answer:

x39x2dxLet u=x3,dv=9x2dx,du=3x2dx,v=x9x2+9arcsin(x3)2=x49x2+9x3arcsin(x3)2+(3x2+18)(9x2)345x3arcsin(x3)9x2(15x2+270)10+C=x49x22+(3x2+18)(9x2)3/210+9x2(3x22+27)+C

WHAT IS GOING ON!!!!

 Apr 13, 2017
 #1
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????      

.
 Apr 13, 2017
 #2
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Take the integral:
 integral x^3 sqrt(9 - x^2) dx

For the integrand x^3 sqrt(9 - x^2), substitute u = x^2 and du = 2 x dx:
 = 1/2 integral sqrt(9 - u) u du
For the integrand sqrt(9 - u) u, substitute s = 9 - u and ds = - du:

= 1/2 integral(s - 9) sqrt(s) ds
Expanding the integrand (s - 9) sqrt(s) gives s^(3/2) - 9 sqrt(s):
 = 1/2 integral(s^(3/2) - 9 sqrt(s)) ds
Integrate the sum term by term and factor out constants:
 = 1/2 integral s^(3/2) ds - 9/2 integral sqrt(s) ds
The integral of s^(3/2) is (2 s^(5/2))/5:
 = s^(5/2)/5 - 9/2 integral sqrt(s) ds
The integral of sqrt(s) is (2 s^(3/2))/3:
 = s^(5/2)/5 - 3 s^(3/2) + constant
Substitute back for s = 9 - u:
 = 1/5 (9 - u)^(5/2) - 3 (9 - u)^(3/2) + constant
Substitute back for u = x^2:
 = 1/5 (9 - x^2)^(5/2) - 3 (9 - x^2)^(3/2) + constant
Which is equal to:
Answer: | = -1/5 (9 - x^2)^(3/2) (x^2 + 6) + constant

 Apr 13, 2017
 #3
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Max: Check your answer against this step-by-step answer here:

http://www.integral-calculator.com/

 Apr 13, 2017

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