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# What the h.e.l.l. is wrong with me?

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I was doing some integrations and one of the problems is: $$\displaystyle \int x^3\sqrt{9-x^2}\mathtt{dx}$$

The answer says: $$-\dfrac{x^2}{3}(9-x^2)^{3/2}-\dfrac{2}{15}(9-x^2)^{5/2}+C$$

$$\qquad\displaystyle \int x^3\sqrt{9-x^2}\mathtt{dx}\\ \text{Let }u=x^3, dv=\sqrt{9-x^2}dx, du = 3x^2dx, v=\dfrac{x\sqrt{9-x^2}+9\arcsin(\dfrac{x}{3})}{2}\\ =\dfrac{x^4\sqrt{9-x^2}+9x^3\arcsin(\dfrac{x}{3})}{2}+\dfrac{(3x^2+18)\sqrt{(9-x^2)^3}-45x^3\arcsin(\dfrac{x}{3})-\sqrt{9-x^2}(15x^2+270)}{10}+C\\ =\dfrac{x^4\sqrt{9-x^2}}{2}+\dfrac{(3x^2+18)(9-x^2)^{3/2}}{10}+\sqrt{9-x^2}(\dfrac{3x^2}{2}+27)+C$$

WHAT IS GOING ON!!!!

MaxWong  Apr 13, 2017
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????

MysticalJaycat  Apr 13, 2017
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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

Guest Apr 13, 2017
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