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# Help with a trig sub

0
1422
5
+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.

Jun 24, 2014

#5
+26329
+11

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)

$$\textstyle{\left {(} \sin{(z)}\cos{(z)} \right{)'}=\cos{(z)}*\cos{(z)}+\sin{(z)}(-\sin{(z)})=\cos^2{(z)}-\sin^2{(z)}}$$

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

$$\textstyle{\left {(} \sin{(z)}\cos{(z)} \right{)'} =cos^2{(z)}-(1-cos^2{(z)}}) =-1+2\cos^2{(z)}$$

$$2\cos^2{(z)}=1+ \left {(} \sin{(z)}\cos{(z)} \right{)'} \quad | \quad \int$$

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

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

Back substitute:

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

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

Jun 25, 2014

#1
+118117
+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

Jun 24, 2014
#2
+576
0

That's pretty nifty. Thanks!

Jun 24, 2014
#3
+33266
+10

Here's the trig substitution approach:

Jun 24, 2014
#4
+576
0

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

Jun 24, 2014
#5
+26329
+11

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)

$$\textstyle{\left {(} \sin{(z)}\cos{(z)} \right{)'}=\cos{(z)}*\cos{(z)}+\sin{(z)}(-\sin{(z)})=\cos^2{(z)}-\sin^2{(z)}}$$

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

$$\textstyle{\left {(} \sin{(z)}\cos{(z)} \right{)'} =cos^2{(z)}-(1-cos^2{(z)}}) =-1+2\cos^2{(z)}$$

$$2\cos^2{(z)}=1+ \left {(} \sin{(z)}\cos{(z)} \right{)'} \quad | \quad \int$$

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

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

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