Integration by trig substitution

[mvantuyl]

Homework Statement

Integrate: \int\sqrt{1-9t^{2}}dt

Homework Equations

The Attempt at a Solution

t = 1/3 sin\Theta
dt/d\Theta = 1/3 cos\Theta
dt = 1/3 cos\Thetad\Theta
3t = sin\Theta

1/3\int\sqrt{1-sin^{2}}\Theta cos\Thetad\Theta
1/3\int cos^{2}\Thetad\Theta
1/3\int(1 + cos 2\Theta) / 2 d\Theta
1/6\int1 + cos2\Theta d\Theta
1/6(\Theta + 1/2 sin 2\Theta) + C
1/6(\Theta + 1/2(sin\Theta cos\Theta) + C
1/6(\Theta + 1/2(sin\Theta\sqrt{1-sin^{2}\Theta})) + C
1/6(\Theta + 1/2(3t\sqrt{1-9t^{2}})) + C

I can't figure out how to get rid of \Theta in the result.

 

[The Bob]

Why not using arcsin? Also, there is a factor of two missing in the 3rd step from the end.

The Bob

 

[mvantuyl]

Perfect! Thanks.