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# Integral

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Integral x times sqrt (0.25 - xsquared) dx

Jun 18, 2015

#9
+117762
+5

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

Jun 20, 2015

#1
+117762
+5

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

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Jun 18, 2015
#2
+26319
+5

Integral x times sqrt (0.25 - xsquared) dx

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

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

Jun 18, 2015
#3
+117762
+5

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

Jun 18, 2015
#4
+26319
+5

Melody,

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

Is here a trick?

Jun 18, 2015
#5
+117762
+5

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

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Jun 18, 2015
#6
+26319
+5

Melody,

Jun 18, 2015
#7
+117762
+5

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 :))

Jun 18, 2015
#8
+26319
+5

Hallo Melody,

thank you very much. This is a good idea.

Jun 19, 2015
#9
+117762
+5

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

Melody Jun 20, 2015