+0  
 
0
424
5
avatar+576 

I'm trying to take an integral bounded below by zero and above by alpha of the sqrt(1-x^2) 

I think It is a trig sub but the answer I keep getting is ridiculous.  Any suggestions are appreciated.

jboy314  Jun 24, 2014

Best Answer 

 #5
avatar+18827 
+10

I'm trying to take an integral bounded below by zero and above by alpha of the sqrt(1-x^2) :

$$\boxed{\int\limits_{x=0}^{x=\alpha } \sqrt{1-x^2}\ dx \quad ?}$$

Substitute x only:   x = sin(z)   and    dx = cos(z) dz

$$\int\limits_{x=0}^{x=\alpha } \underbrace{\sqrt{1-\sin^2{(z)}}}_{\cos{(z)}} \overbrace{\cos{(z)} \ dz }^{\ dx}=\int\limits_{x=0}^{x=\alpha }\cos^2{(z)} \ dz$$

Product rule (uv)' = u'v+uv'

u =sin(z)     v =cos(z)

u'=cos(z)     v'=-sin(z)

$$\textstyle{\left \textcolor[rgb]{1,0,0}{(} \sin{(z)}\cos{(z)} \right\textcolor[rgb]{1,0,0}{)'}=\cos{(z)}*\cos{(z)}+\sin{(z)}(-\sin{(z)})=\cos^2{(z)}-\sin^2{(z)}}$$

$$\boxed{\sin^2{(z)}=1-\cos^2{(z)} }$$

$$\textstyle{\left \textcolor[rgb]{1,0,0}{(} \sin{(z)}\cos{(z)} \right\textcolor[rgb]{1,0,0}{)'}
=cos^2{(z)}-(1-cos^2{(z)}}) =-1+2\cos^2{(z)}$$

$$2\cos^2{(z)}=1+ \left \textcolor[rgb]{1,0,0}{(} \sin{(z)}\cos{(z)} \right\textcolor[rgb]{1,0,0}{)'} \quad | \quad \int$$

$$2\int{\cos^2{(z)}}\ dz
=
\underbrace{\int{\ dz }}_z
+ \sin{(z)}\cos{(z)}$$

$$\boxed{\int{\cos^2{(z)}}\ dz
=\dfrac{1}{2}
\left(
z+\sin{(z)}\cos{(z)}
\right)}$$

Back substitute:

$$\\z=\sin^{-1}{(x)}\\
\sin{(z)}=x\\
\cos{(z)}=\sqrt{1-x^2}$$

$$\begin{array}{rcl}
\int\limits_{x=0}^{x=\alpha } \sqrt{1-x^2}\ dx &=&
\left[
\dfrac{1}{2}
\left(
\sin^{-1}{(x)}+x\sqrt{1-x^2}
\right)
}
\right]^{x=\alpha}_{x=0}\\\\
&=&\frac{1}{2}
\left(\;
\sin^{-1}{(\alpha)}+\alpha\sqrt{1-\alpha^2}
\;\right)


\end{array}$$

heureka  Jun 25, 2014
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5+0 Answers

 #1
avatar+91412 
+5

I don't know but I found this on the web.

http://math.stackexchange.com/questions/533082/integral-of-sqrt1-x2-using-integration-by-parts

Maybe it will help you?

Melody  Jun 24, 2014
 #2
avatar+576 
0

That's pretty nifty. Thanks!

jboy314  Jun 24, 2014
 #3
avatar+26397 
+10

Here's the trig substitution approach:

integral

Alan  Jun 24, 2014
 #4
avatar+576 
0

Looks like i totally blew the first step in my own work by not using a substitution for dx!  wow!  Thanks!

jboy314  Jun 24, 2014
 #5
avatar+18827 
+10
Best Answer

I'm trying to take an integral bounded below by zero and above by alpha of the sqrt(1-x^2) :

$$\boxed{\int\limits_{x=0}^{x=\alpha } \sqrt{1-x^2}\ dx \quad ?}$$

Substitute x only:   x = sin(z)   and    dx = cos(z) dz

$$\int\limits_{x=0}^{x=\alpha } \underbrace{\sqrt{1-\sin^2{(z)}}}_{\cos{(z)}} \overbrace{\cos{(z)} \ dz }^{\ dx}=\int\limits_{x=0}^{x=\alpha }\cos^2{(z)} \ dz$$

Product rule (uv)' = u'v+uv'

u =sin(z)     v =cos(z)

u'=cos(z)     v'=-sin(z)

$$\textstyle{\left \textcolor[rgb]{1,0,0}{(} \sin{(z)}\cos{(z)} \right\textcolor[rgb]{1,0,0}{)'}=\cos{(z)}*\cos{(z)}+\sin{(z)}(-\sin{(z)})=\cos^2{(z)}-\sin^2{(z)}}$$

$$\boxed{\sin^2{(z)}=1-\cos^2{(z)} }$$

$$\textstyle{\left \textcolor[rgb]{1,0,0}{(} \sin{(z)}\cos{(z)} \right\textcolor[rgb]{1,0,0}{)'}
=cos^2{(z)}-(1-cos^2{(z)}}) =-1+2\cos^2{(z)}$$

$$2\cos^2{(z)}=1+ \left \textcolor[rgb]{1,0,0}{(} \sin{(z)}\cos{(z)} \right\textcolor[rgb]{1,0,0}{)'} \quad | \quad \int$$

$$2\int{\cos^2{(z)}}\ dz
=
\underbrace{\int{\ dz }}_z
+ \sin{(z)}\cos{(z)}$$

$$\boxed{\int{\cos^2{(z)}}\ dz
=\dfrac{1}{2}
\left(
z+\sin{(z)}\cos{(z)}
\right)}$$

Back substitute:

$$\\z=\sin^{-1}{(x)}\\
\sin{(z)}=x\\
\cos{(z)}=\sqrt{1-x^2}$$

$$\begin{array}{rcl}
\int\limits_{x=0}^{x=\alpha } \sqrt{1-x^2}\ dx &=&
\left[
\dfrac{1}{2}
\left(
\sin^{-1}{(x)}+x\sqrt{1-x^2}
\right)
}
\right]^{x=\alpha}_{x=0}\\\\
&=&\frac{1}{2}
\left(\;
\sin^{-1}{(\alpha)}+\alpha\sqrt{1-\alpha^2}
\;\right)


\end{array}$$

heureka  Jun 25, 2014

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