- #1
Sink41
- 21
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EDIT: my tex is a little broken trying to fix
So i want to intergrate
[tex]
\int (1 - x^2)^\frac{1}{2} dx
[/tex]
i start off by saying
[tex]\sin u = x[/tex]
so
[tex]
\frac {dx} {du} = \cos u
[/tex]
then
[tex]
\int (1 - x^2)^\frac{1}{2} \cos u du
[/tex]
which is
[tex]
\int \cos^2 u du
[/tex]
and
[tex]
\cos2u = 2cos^2 u - 1
[/tex]
so therefore
[tex]
\frac {1} {2} \cos2u + \frac {1} {2}= cos^2 u
[/tex]
so you intergrate
[tex]
\int \frac {1} {2} \cos 2u + \frac {1} {2} du
[/tex]
which is
[tex]
\frac {1} {4} \sin 2u + \frac {u} {2}
[/tex]
and
[tex]
\sin 2u = 2\sin u \cos u = 2\sin u(1 - sin^2 u)^\frac {1} {2}
[/tex]
putting x into u gets
[tex]
\frac {x} {2} (1 - x^2)^\frac {1} {2} + \frac {\sin^-1 x} {2}
[/tex]
which I am pretty sure is wrong. So can someone show me how to intergrate (1-(X^2))^0.5 ? i think that using x = sinu is wrong but u = sinx doesn't get me far either. Probably something simple I've overlooked.
So i want to intergrate
[tex]
\int (1 - x^2)^\frac{1}{2} dx
[/tex]
i start off by saying
[tex]\sin u = x[/tex]
so
[tex]
\frac {dx} {du} = \cos u
[/tex]
then
[tex]
\int (1 - x^2)^\frac{1}{2} \cos u du
[/tex]
which is
[tex]
\int \cos^2 u du
[/tex]
and
[tex]
\cos2u = 2cos^2 u - 1
[/tex]
so therefore
[tex]
\frac {1} {2} \cos2u + \frac {1} {2}= cos^2 u
[/tex]
so you intergrate
[tex]
\int \frac {1} {2} \cos 2u + \frac {1} {2} du
[/tex]
which is
[tex]
\frac {1} {4} \sin 2u + \frac {u} {2}
[/tex]
and
[tex]
\sin 2u = 2\sin u \cos u = 2\sin u(1 - sin^2 u)^\frac {1} {2}
[/tex]
putting x into u gets
[tex]
\frac {x} {2} (1 - x^2)^\frac {1} {2} + \frac {\sin^-1 x} {2}
[/tex]
which I am pretty sure is wrong. So can someone show me how to intergrate (1-(X^2))^0.5 ? i think that using x = sinu is wrong but u = sinx doesn't get me far either. Probably something simple I've overlooked.
Last edited: