How to integrate without trigonometric substitution I need to integrate the following equation without using trigonometric substitution (which we haven't learned yet but I have been told would be the normal way to integrate it).
$\int_0^2 (y+8)\sqrt {-y^2+4y}$ $dx$
I know that the answer to this is
$ 10\pi−\frac{8}{3} $ but I don't know how to demonstrate this.
I know that it's useful that the integral of $\sqrt {-y^2+4y}$ is just [a fourth of] the area of a circle with radius 2 but unless I can isolate the radical from (y+8) I can't use this fact. Thoughts?
 A: Hint:
$$(y+8)\sqrt{-y^2+4y}$$
$$=(-\frac{1}{2}(-2y)+8)\sqrt{-y^2+4y}$$
$$=(-\frac{1}{2}(-2y+4)+10) \sqrt{-y^2+4y}$$
$$=-\frac{1}{2}(-2y+4)\sqrt{-y^2+4y}+10 \sqrt{-y^2+4y}$$
$\int_{0}^{2} \sqrt{-y^2+4y} \, dy$ represents a fourth of the area of a circle of radius $2$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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Use Euler Substitution $\ds{y = {\ic t^{2} \over 2\pars{t + 2\ic}}}$
  with $\ds{t = \root{-y^{2} + 4y} - y\ic}$. You'll get

\begin{align}
&\int_{0}^{2}\pars{y + 8}\root{-y^{2} + 4y}\,\dd y
\\[5mm] = &\
\int_{0}^{2 - 2\ic}\bracks{-5 + 2\ic t - {t^{2} \over 8} +
{8 \over \pars{t + 2\ic}^{4}} + {40\ic \over \pars{t + 2\ic}^{3}} +
{2 \over \pars{t + 2\ic}^{2}} +
{20\ic \over t + 2\ic}}\,\dd t
\end{align}
which involves straightforward integrations.
The answer is $\bbx{\ds{10\pi - {8 \over 3}}}$.
A: Express $y+8$as a differential of $-y^2+4y$ and some constant.Then take $-y^2+4y$ as $t$ and proceed.what i meant was write $y+8=a(\frac{d}{dx} [-y^2+4]+b$.Then solve for a and b and put in the integral.
