Finding the final form of ∫√(1−9t²)dt 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$$\Theta$$d$$\Theta$$
3t = sin$$\Theta$$

1/3$$\int\sqrt{1-sin^{2}}\Theta$$ cos$$\Theta$$d$$\Theta$$
1/3$$\int cos^{2}\Theta$$d$$\Theta$$
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.