Find area bounded by functions $y_1=\sqrt{4x-x^2}$ and $y_2=x\sqrt{4x-x^2}$. From $y_1=y_2\Rightarrow x=1$. Intersection points of $y_1$ and $y_2$ are $A(0,0),B(1,\sqrt 3),C(4,0)$. Domain of $y_1$ and $y_2$ is $x\in [0,4]$. On the interval $x\in[0,1]\Rightarrow y_1\ge y_2$ and on the interval $x\in[1,4]\Rightarrow y_1\le y_2$.
$$A=\int_0^1 (y_1-y_2)\mathrm dx+\int_1^4 (y_2-y_1)\mathrm dx=\int_0^1 (1-x)\sqrt{4x-x^2}\mathrm dx+\int_1^4 (x-1)\sqrt{4x-x^2}\mathrm dx$$
How to solve integrals $\int  \sqrt{4x-x^2}\mathrm dx$ and $\int  x\sqrt{4x-x^2}\mathrm dx$?
Substitution $$u=\sqrt{\frac{x}{4-x}}\Rightarrow du=\frac{2}{(x-4)^2\sqrt{\frac{x}{4-x}}}dx$$ doesn't seems to work.
 A: If you'd like to get rid of radicals, you can do the following substitution:
$$z=\sqrt{\frac{4}{x}-1}$$
$$x=\frac{4}{1+z^2}$$
$$dx=-\frac{8z}{(1+z^2)^2}dz$$
The integrals will become:
$$I_1=\int \sqrt{4x-x^2}dx=-32\int \frac{z^2}{(1+z^2)^3}dz$$
$$I_2=\int x \sqrt{4x-x^2}dx=-128\int \frac{z^2}{(1+z^2)^4}dz$$

$$I_1=32\left(\int \frac{1}{(1+z^2)^3}dz -\int \frac{1}{(1+z^2)^2}dz \right)$$
$$I_2=128\left(\int \frac{1}{(1+z^2)^4}dz -\int \frac{1}{(1+z^2)^3}dz \right)$$
These integrals contain only rational functions. Moreover, we know the integral:
$$\int \frac{1}{1+z^2}dz=\arctan(z)+C$$
Note that:
$$z(0)=\infty$$
$$z(1)=\sqrt{3}$$
$$z(4)=0$$

I don't know what methods you are supposed to use, but I myself would use a parameter. (I will omit the constants of integration).
$$I_0=\int \frac{1}{1+a^2z^2}dz=\frac{\arctan(az)}{a}$$
$$\frac{\partial I_0}{\partial a}= -\int \frac{2az^2}{(1+a^2z^2)^2}dz=\frac{z}{a(1+a^2z^2)}-\frac{\arctan(az)}{a^2}$$
$$a \to 1$$
$$\int \frac{2z^2}{(1+z^2)^2}dz=\arctan(z)-\frac{z}{1+z^2}$$
$$\int \frac{1}{1+z^2}dz-\int \frac{1}{(1+z^2)^2}dz=\frac{1}{2} \left( \arctan(z)-\frac{z}{1+z^2} \right)$$

$$\int \frac{1}{(1+z^2)^2}dz=\frac{1}{2} \left( \arctan(z)+\frac{z}{1+z^2} \right)$$
The same way you can calculate the rest of the itegrals.
