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.
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:
$$\\z=\sin^{-1}{(x)}\\
\sin{(z)}=x\\
\cos{(z)}=\sqrt{1-x^2}$$
$$\begin{array}{rcl}
\int\limits_{x=0}^{x=\alpha } \sqrt{1-x^2}\ dx &=&
\left[
\dfrac{1}{2}
\left(
\sin^{-1}{(x)}+x\sqrt{1-x^2}
\right)
}
\right]^{x=\alpha}_{x=0}\\\\
&=&\frac{1}{2}
\left(\;
\sin^{-1}{(\alpha)}+\alpha\sqrt{1-\alpha^2}
\;\right)
\end{array}$$
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?
Looks like i totally blew the first step in my own work by not using a substitution for dx! wow! Thanks!
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:
$$\\z=\sin^{-1}{(x)}\\
\sin{(z)}=x\\
\cos{(z)}=\sqrt{1-x^2}$$
$$\begin{array}{rcl}
\int\limits_{x=0}^{x=\alpha } \sqrt{1-x^2}\ dx &=&
\left[
\dfrac{1}{2}
\left(
\sin^{-1}{(x)}+x\sqrt{1-x^2}
\right)
}
\right]^{x=\alpha}_{x=0}\\\\
&=&\frac{1}{2}
\left(\;
\sin^{-1}{(\alpha)}+\alpha\sqrt{1-\alpha^2}
\;\right)
\end{array}$$