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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 im 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 doesnt get me far either. Probably something simple ive 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 im 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 doesnt get me far either. Probably something simple ive overlooked.

Last edited: