<s>[INST] Using trigonometric substitution to integrate $\int\frac{x^3dx}{\sqrt{25-x^2}}$ 
I tried solving this calculus problem so many times but I never got an answer that at least looked similar to the choices in the image. Can someone help?
I don't understand where the Sine came from. This is the answer I get
$$\frac13(25-x^2)^{3/2} - 25\sqrt{25-x^2} + C $$

Edits are appreciated
 [/INST] Integration by parts. Your result is equal to mine, BTW
$$\int \frac{x^3}{\sqrt{25-x^2}} \, dx=\int (-x^2)\,\frac{ -x}{\sqrt{25-x^2}}\,dx=$$
$$\int \frac{ -x}{\sqrt{25-x^2}}\,dx=\sqrt{25-x^2}+C$$
$$=-x^2\sqrt{25-x^2}-\int\left(-2x\sqrt{25-x^2}\right)\,dx=$$
$$=-x^2\sqrt{25-x^2}-\frac{2}{3} \left(25-x^2\right)^{3/2}+C=$$
$$=-x^2\sqrt{25-x^2}-\frac{2}{3}\sqrt{25-x^2}(25-x^2)+C=$$
$$=-\sqrt{25-x^2}\left(x^2+\frac{2}{3}(25-x^2)\right)+C=-\frac{1}{3} \sqrt{25-x^2} \left(x^2+50\right)+C$$
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