Integral

I am considering the very elegant substitution that Alan used here

http://web2.0calc.com/questions/integrate-cos-1x-2-1-x-2-1-2

I am going to look for something like that here.

Heureka used a substitution but it looked hard to me.  

$$\\\int x\;\sqrt {0.25 - x^2} dx\\\\
Let \\
u=0.25-x^2\\\\
du=-2x\;dx\\\\
dx=\frac{du}{-2x}\\\\
so\\
\int x\;\sqrt {0.25 - x^2} dx\\\\
=\int \not{x}\;\sqrt {u}\; \frac{du}{-2\not{x}}\\\\
=\int \;\frac{u^{1/2}}{-2}\; du\\\\
=\frac{2u^{3/2}}{3*-2}+c\\\\
=\frac{-u^{3/2}}{3}+c\\\\
=\frac{-(0.25-x^2)^{3/2}}{3}+c\\\\$$

YES I Like that - thanks Alan 

$$\\\int x\sqrt{0.25-x^2}\;dx\\\\
=\int x(0.25-x^2)^{1/2}\;dx\\\\
=(0.25-x^2)^{3/2}\times \frac{2}{3}\times\frac{-1}{2}+c\\\\
=(0.25-x^2)^{3/2}\times \frac{2}{3}\times\frac{-1}{2}+c\\\\
=(0.25-x^2)^{3/2}\times \frac{-1}{3}+c\\\\
=\frac{-(0.25-x^2)^{3/2}}{3}+c\\\\$$

Integral x times sqrt (0.25 - xsquared) dx

$$\small{\text{$
\begin{array}{lrcl}
& \int{x\sqrt{\frac{1}{4}-x^2 }} \ dx &=& \frac{1}{2}\int{x \sqrt{1-(2x)^2} \ dx } \\\\
\mathrm{We~ substitute:~} & 2x &=& \sin{(z)} \\
&2\ dx &=& \cos{(z)}\ dz \\\\
& \int{x\sqrt{\frac{1}{4}-x^2 }} \ dx &=& \frac{1}{2}\int{
\frac{\sin{(z)} }{2}
\sqrt{1-[\sin{(z)} ]^2} \cdot \frac{ \cos{(z)} }{2}\ dz } \\\\
& \int{x\sqrt{\frac{1}{4}-x^2 }} \ dx &=&
\frac{1}{8} \int{
\sin{(z)} \cdot \cos{(z)} \cdot \cos{(z)} \ dz }\\\\
& \int{x\sqrt{\frac{1}{4}-x^2 }} \ dx &=&
\frac{1}{8} \int{
\sin{(z)} \cdot \cos^2{(z)}\ dz }\\\\
\end{array}
$}}\\\\
\boxed{
\begin{array}{lrcl}
\mathrm{Formula:~} & y &=& \frac{ \cos^{n+1}{(z)} } {n+1} \\ \\
& y' &=& \frac{ (n+1)\cdot \cos^{n+1-1}{(z)} \cdot [-\sin{(z)}] }{n+1} \\\\
& y' &=& -\sin{(z)}\cdot \cos^{n}{(z)} \\\\
\mathrm{so:~} & -\int{ \sin{(z)}\cdot \cos^{n}{(z)} \ dz }
&=& \frac{ \cos^{n+1}{(z)} } {n+1} \\\\
& \mathbf{\int{ \sin{(z)}\cdot \cos^{n}{(z)} \ dz } }
&\mathbf{=}& \mathbf{ -\frac{ \cos^{n+1}{(z)} } {n+1} } \\\\
& \mathbf{\int{ \sin{(z)}\cdot \cos^{2}{(z)} \ dz } }
&\mathbf{=}& \mathbf{ -\frac{ \cos^{3}{(z)} } {3} }
\end{array}
}$$

$$\small{\text{$
\begin{array}{lrcl}
& \int{x\sqrt{\frac{1}{4}-x^2 }} \ dx &=&
\frac{1}{8} \int{
\sin{(z)} \cdot \cos^2{(z)}\ dz }\\\\
& \int{x\sqrt{\frac{1}{4}-x^2 }} \ dx &=&
\frac{1}{8} \left[-\frac{ \cos^{3}{(z)} } {3} \right] + c \\\\
& \int{x\sqrt{\frac{1}{4}-x^2 }} \ dx &=&
-\frac{1}{24} \cdot \cos^{3}{(z)} + c \qquad | \qquad \cos{(z)} = \sqrt{1-(2x)^2}\\\\
& \int{x\sqrt{\frac{1}{4}-x^2 }} \ dx &=&
-\frac{1}{24} \cdot \left(\sqrt{1-(2x)^2} \right)^3 + c\\\\
& \int{x\sqrt{\frac{1}{4}-x^2 }} \ dx &=&
-\frac{1}{24} \cdot \left(1-4x^2\right)^{\frac32} + c
\end{array}
$}}$$

Heureka, do you think my answer is wrong?  They are probably both  the same 

Melody,

i have seen, they are probably both  the same. But i have not understood your way.

Is here a trick?

$$\\\frac{-(0.25-x^2)^{3/2}}{3}+c\\\\
=\frac{-(\frac{1-4x^2}{4})^{3/2}}{3}+c\\\\
=\frac{-\left(\frac{(1-4x^2)^{3/2}}{4^{3/2}}\right)}{3}+c\\\\
=-\left(\frac{(1-4x^2)^{3/2}}{8}\right)\times \frac{1}{3}+c\\\\
=-\frac{(1-4x^2)^{3/2}}{24}+c\\\\
=-\frac{1}{24}\;(1-4x^2)^{3/2}+c\\\\
=\;$Heureka's answer :)$$$

Melody,

your answer is the same, but how you have integrated ?

No, it wasn't a trick Heureka

$$\int x\sqrt{0.25-x^2}\;dx\\\\
=\int x(0.25-x^2)^{1/2}\;dx\\\\
$Now I noted that $\frac{d}{dx}(0.25-x^2)=-2x \;\;$So the answer is going to be a multiple of $(0.25-x^2)^{1.5}\\\\\\
\frac{d}{dx}(0.25-x^2)^{1.5}\\\\
=\frac{3}{2}(0.25-x^2)^{0.5}*-2x\\\\
=\frac{3}{2}*-2x(0.25-x^2)^{0.5}\\\\\\
$so going backwards$\\\\
=\int \frac{3}{2}*-2x(0.25-x^2)^{1.5}dx\\\\
=x\sqrt{0.25-x^2}+c\\\\$$

$$\\so\\\\
\int \frac{3}{2}*-2x(0.25-x^2)^{1.5}dx=x(0.25-x^2)^{0.5}+c\\\\
\int \frac{2}{3}*\frac{-1}{2}*\frac{3}{2}*-2x(0.25-x^2)^{1.5}dx=\frac{2}{3}*\frac{-1}{2}*(0.25-x^2)^{0.5}+c\\\\
\int x(0.25-x^2)^{1.5}dx=\frac{2}{3}*\frac{-1}{2}*(0.25-x^2)^{0.5}+c\\\\
\int x(0.25-x^2)^{1.5}dx=\frac{-1}{3}*(0.25-x^2)^{0.5}+c\\\\$$

I didn't need to do all that.  i just took the reciprocal of the -2 and the 3/2 in the first place

$$\frac{-1}{2}\times \frac{2}{3}=-\frac{1}{3}$$

Sorry I took so long I was having a hard time explainining.  

I kind of 'stumble' my way through these.  It was good that you made me think about it properly   

(I hope I didn't stuff up )

If i had to do it your way I'd never have gotten it out!  

It is usually me that has the reputation for doing it the LONG way.  Looks like it was your turn :))

Hallo Melody,

thank you very much. This is a good idea.