# Integration by trig substitution

1. Mar 5, 2009

### mvantuyl

1. The problem statement, all variables and given/known data
Integrate: $$\int$$$$\sqrt{1-9t^{2}}$$dt

2. Relevant equations

3. 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.

2. Mar 5, 2009

### The Bob

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

The Bob

3. Mar 5, 2009

### mvantuyl

Perfect! Thanks.