Problem understanding integration by parts from the book An example from my book (from unrelated subject) has the following steps:
$$ -2 \int \sqrt{9-y^2} dy = - y\sqrt{9-y^2} + 9\sin^{-1}\frac{y}{3} \  +  C$$
I am trying to understand the steps in between from my books example when they use the integral forumla (second image below).
I understand that this is integration by parts and they use the integral formula on the bottom. Apparently, they set the parts to $$ u(y) = \sqrt{9-y^2} \\ v(y) = y $$
And thats how the term $y\sqrt{9-y^2}$ is calculated but I dont understand how they do with the next term in the partial integration. My attempt:
$$ - \int^3_0 \left(\sqrt{9-y^2}\right)' y \ dy = - \int^3_0 \frac{1}{2\sqrt{9-y^2}} \cdot 2y \cdot y \ dy = \dots $$
The term that popped up is $2y$ which is the result of derivative of the inner function $9-y^2$. This result does not seem to be right. So what steps do they take here?




As mentioned by @Deepak you can use substitution and find the answer that way.
Next it should be
$$-2\int \sqrt{9-y^2}dy=-y\sqrt{9-y^2}\color{red}{-}9\sin^{-1}\big(\frac{y}{3}\big)+C$$
Here is the integration by parts method
First note that $\frac{d}{dy}\big(\sqrt{9-y^2}\big)=\color{red}{-}\frac{y}{\sqrt{9-y^2}},$ then we have
$$\int \sqrt{9-y^2}dy=\int u(y)v'(y)dy=u(y)v(y)-\int u'(y)v(y)dy$$
$$=y\sqrt{9-y^2}\color{blue}{+}\int \frac{y^2}{\sqrt{9-y^2}}dy$$
$$=y\sqrt{9-y^2}\color{blue}-\int \frac{-y^2+\color{red}{9}-\color{red}{9}}{\sqrt{9-y^2}}dy$$
$$=y\sqrt{9-y^2}-\int\frac{9-y^2}{\sqrt{9-y^2}}+9\int\frac{1}{\sqrt{9-y^2}}dy$$
$$=y\sqrt{9-y^2}-\int\sqrt{9-y^2}dy+9\sin^{-1}\big(\frac{y}{3}\big)-C$$
and rearranging gives
$$2\int{\sqrt{9-y^2}}dy=y\sqrt{9-y^2}+9\sin^{-1}\big(\frac{y}{3}\big)-C$$
or $$-2\int{\sqrt{9-y^2}}dy=-y\sqrt{9-y^2}-9\sin^{-1}\big(\frac{y}{3}\big)+C$$
