Krdytkyu

Recently I've been working on problems including integration and I'm having some difficulties solving this integral :

$$int_0^1 (1-x)sqrt{4x-x^2} {dx}$$

I tried to rewrite it in this form:

$$int_0^1 (1-x)xsqrt{frac{4-x}{x}} dx$$ and then subsitute $frac{4-x}{x}=t^2$ and by this substitution I get a function way easier integrated of this form:

$$-32intfrac{(t^2-3)t^2}{(1+t^2)^4}dt$$

I can integrate this function easily by simplifying it but the problem is that I cannot define its borders because $t$ is not defined for $x=0$ so this means that my substitution doesn't hold. But I don't have any other idea how to solve it so any help will be appreciated.Thank you!

It is always didactically good to give a solution for the given start. I will try it, although trigonometric substitutions from the beginning may save typing.

The substitution $t=sqrt{(4-x)/x}$ is ok, but we have to use improper integrals. The solution would be as follows. For $x=0+$ the corresponding (limiting) value of $t$ is $t=sqrt{4/0_+}=+infty$, and for $x=1$ we get $t=sqrt{(4-1)/1}=color{red}{sqrt 3}$. (It is square three, maybe the one reason for writing this answer.)
$$
begin{aligned}
J &=
int_0^1 (1-x)sqrt{4x-x^2} ; dx
\
&=
int_{infty}^{sqrt 3}
-32frac{(t^2-3)t^2}{(1+t^2)^4};dt
\
&=
+32int_{sqrt 3}^{infty}
frac{(t^2+2t^2+1)-5(t^2+1)+4}{(1+t^2)^4};dt
\
&=
+32int_{sqrt 3}^{infty}
left[
frac1{(1+t^2)^2}
-5frac1{(1+t^2)^3}
+4frac1{(1+t^2)^4}
right];dt
\
&=32(K_2-5K_3+4K_4) ,
end{aligned}
$$
where $K_n$ is the integral on the same interval of $(1+t^2)^{-n}$. We have the recursion
$$
begin{aligned}
K_n
&=int_{sqrt 3}^{infty}t'frac1{(1+t^2)^n};dt
\
&=frac t{(1+t^2)^n}Bigg|_{sqrt 3}^{infty}
-
int_{sqrt 3}^{infty}
tcdot frac{-2nt}{(1+t^2)^{n+1}};dt
\
&=-frac{sqrt3}{4^n}
+
2nint_{sqrt 3}^{infty}frac{(t^2+1)-1}{(1+t^2)^{n+1}};dt
\
&=-frac{sqrt3}{4^n}
+
2n(K_n-K_{n+1}) ,text{ i.e.}
\
K_{n+1}
&=
frac 1{2n}left[ (2n-1)K_n-frac{sqrt3}{4^n} right] .
\[2mm]
&qquadtext{This gives:}
\
K_1 &=arctan infty-arctansqrt 3 =left(frac 12-frac 13right)pi
=frac 16pi ,
\
K_2 &=frac 12left[K_1-frac{sqrt 3}4right]=frac 1{12}pi - frac{sqrt 3}8 ,
\
K_3 &=frac 14left[3K_2-frac{sqrt 3}{16}right]
=frac 1{16}pi - frac{7sqrt 3}{64} ,
\
K_4 &=frac 16left[5K_3-frac{sqrt 3}{64}right]
=frac 5{96}pi - frac{3sqrt 3}{32} ,
%\
%K_5 &=frac 18left[7K_4-frac{sqrt 3}{216}right]
%=frac {35}{768}pi - frac{169sqrt 3}{2048} ,
\
&qquadtext{ so putting all together}
\
J&=32(K_2-5K_3+4K_4)
\
&=-frac23pi+frac {3sqrt 3}2 .
end{aligned}
$$
As said, not the quick way, but all details are displayed. Computer check:

sage: var('x,t');

sage: integral( (1-x)*sqrt(4*x-x^2), x, 0, 1 ).simplify()

-2/3*pi + 3/2*sqrt(3)

sage: integral( 32*(t^2-3)*t^2/(1+t^2)^4, t, sqrt(3), oo ).simplify()

-2/3*pi + 3/2*sqrt(3)

The us try to get rid of the square root step-by-step:
$$begin{eqnarray*} int_{0}^{1}(1-x)sqrt{x}sqrt{4-x},dx&stackrel{xmapsto z^2}{=}& 2int_{0}^{1}z^2(1-z^2)sqrt{4-z^2},dz\&stackrel{zmapsto 2u}{=}&32int_{0}^{1/2}u^2(1-4u^2)sqrt{1-u^2},du\&stackrel{umapstosintheta}{=}&32int_{0}^{pi/6}sin^2thetacos^2theta(1-4sin^2theta),dtheta\&=&4int_{0}^{pi/6}left[-1+cos(2theta)+cos(4theta)-cos(6theta)right],dtheta.end{eqnarray*}$$
I guess you may take it from here.

One can use $,displaystyle{int_a^b f(x) dx=int_a^b f(a+b-x) dx},$ to get: $$I=int_0^1 x sqrt{4-(1+x)^2} dx= int_0^1 (x+1)sqrt{4 - (x+1)^2} dx-int_0^1 sqrt{4 - (x+1)^2} dx$$ For the first one just substitute $4-(x+1)^2 =t$ to get $displaystyle{frac12 int_0^3 sqrt t dt}., $ And the second one is a standard square root integral.

Hint:

By completing the square,

$$4x-x^2=4-(x-2)^2$$ and that hints to use the substitution

$$x-2=2cos t.$$

Then

$$int (1-x)sqrt{4x-x^2}dx=4int(1+2cos t)sin^2t,dt.$$

The term $sin^2t$ is integrated in the form $dfrac{1-cos 2t}2$, and the second term is immediate.

$$I=2t-2sin tcos t+dfrac83sin^3t=\arccosdfrac{x-2}2-(x-2)sqrt{1-dfrac{(x-2)^2}4}+dfrac83left(1-dfrac{(x-2)^2}4right)^{3/2}$$

Your approach can be completed by noting that the domain of integrtion $xin(0,1)$ gets transformed to the unbounded domain of integration $tin(sqrt3,infty)$ with reversal of orientation (which negates the integral).

We can also set $u=1-x/2implies4!left(1-u^2right)=4x-x^2$. Therefore,
$$
int_0^1(1-x)sqrt{4x-x^2},mathrm{d}x=4int_{1/2}^1(2u-1)sqrt{1-u^2},mathrm{d}u
$$
Then, setting $u=sin(theta)$ gives
$$
begin{align}
4int_{1/2}^1(2u-1)sqrt{1-u^2},mathrm{d}u
&=4int_{pi/6}^{pi/2}(2sin(theta)-1)overbrace{cos^2(theta)}^{1-sin^2(theta)},mathrm{d}theta\
&=4int_{pi/6}^{pi/2}left(-2color{#C00}{sin^3(theta)}+color{#090}{sin^2(theta)}+2sin(theta)-1right)mathrm{d}theta\
&=4int_{pi/6}^{pi/2}left(-2color{#C00}{frac{3sin(theta)-sin(3theta)}4}+color{#090}{frac{1-cos(2theta)}2}+2sin(theta)-1right)mathrm{d}theta\
%&=-left[frac23cos(3theta)+sin(2theta)+2cos(theta)+2thetaright]_{pi/6}^{pi/2}\[3pt]
%&=frac32sqrt3-frac23pi
end{align}
$$