I was doing some integrations and one of the problems is: ∫x3√9−x2dx
The answer says: −x23(9−x2)3/2−215(9−x2)5/2+C
Here is my answer:
∫x3√9−x2dxLet u=x3,dv=√9−x2dx,du=3x2dx,v=x√9−x2+9arcsin(x3)2=x4√9−x2+9x3arcsin(x3)2+(3x2+18)√(9−x2)3−45x3arcsin(x3)−√9−x2(15x2+270)10+C=x4√9−x22+(3x2+18)(9−x2)3/210+√9−x2(3x22+27)+C
WHAT IS GOING ON!!!!
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