<s>[INST] Integration by parts, error on Primitive of $\sqrt{1-x^{2}}$ Problem: Integrate $\sqrt{1-x^{2}}$
Attempt: 
 [/INST] The usual method uses the substitution $x=\sin\theta$ followed by the identity $\cos^2\theta=\frac{1}{2}(\cos 2\theta+1)$. However, since the OP used integration by parts, let's do it that way.  It is perhaps a little harder, but introduces a useful technique.
Let our integral be $I$, and let $g'(x)=1$ and $f(x)=\sqrt{1-x^2}$. Then we get 
$$I=x\sqrt{1-x^2}+\int \frac{x^2}{\sqrt{1-x^2}}\,dx.$$
Rewrite $\frac{x^2}{\sqrt{1-x^2}}$ as $\frac{1-(1-x^2)}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}-\sqrt{1-x^2}$. We get
$$I=x\sqrt{1-x^2} +\int \frac{1}{\sqrt{1-x^2}}\,dx-I.$$
But $\frac{1}{\sqrt{1-x^2}}$ has $\arcsin x$ as an antiderivative. Bring the second $I$ to the other side, divide by $2$, and don't forget about the $+C$. 
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