Evaluating $\int{x^2\over{\sqrt{2x-x^2}}}dx=$ using trig substitution: $(x-1)=\sin\theta$
$\int{x^2\over{\sqrt{2x-x^2}}}dx=I$
$I= \int{{\left(\sin\theta-1\right)^2}\over{\cos\theta}}{\cos\theta d\theta}=$ $\int{{(\sin\theta-1)^2} d\theta}$
$\implies I={\int({\sin^2\theta-2\sin\theta+1 )d\theta}}$
$\implies I={\frac 12}{\int1-\cos2\theta d\theta-\int2\sin\theta d\theta+\int1 d\theta}$
$\implies I={\frac 32\theta}-{\frac 14 \sin2\theta}+{2\cos\theta}+C$
Now,
${\theta=\sin^{-1}(x-1)}$, ${\sin\theta=(x-1)}$ and ${\cos\theta={\sqrt{(2x-x^2)}}}$
We get:
$I={\frac 32\theta}-{\frac 14 \sin2\theta}+{2\cos\theta}+C=$ ${\frac 32 \sin^{-1}(x-1)}-{\frac 12 \sin\theta \cos\theta}+{2\cos\theta}+C$
$\implies I={\frac 32 \sin^{-1}(x-1)}-{\frac 12 (x-1){\sqrt{(2x-x^2)}}}+{2{\sqrt{(2x-x^2)}}}+C$
My answer doesn't add up to the answer written in my textbook.
${\frac 32 \sin^{-1}(x-1)}-{\frac 12 (x-1){\sqrt{(2x-x^2)}}}-{2{\sqrt{(2x-x^2)}}}+C$
Could someone explain how do we get the negative ${-2{\sqrt{(2x-x^2)}}}$ ?
There is an error in the third line, it's $I = \int (\sin \theta + 1)^2$ instead of $\int (\sin \theta -1)^2$.
You probably substituted $x = \sin \theta -1$ instead of $x=\sin\theta +1$ in the numerator.