# 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 x
dt/dx = 1/3 cos x
dt = 1/3 cos x dx
3t = sin x

1/3 $$\int\sqrt{1 - sin^{2} x}$$ cos x dx
1/3 $$\int cos ^{2}x dx$$
1/3 $$\int(1 + cos 2x) / 2$$ dx
1/6 $$\int$$1 + cos 2x dx
1/6 (x + 1/2 sin 2x) + C
1/6 (x + 1/2 (sin x cos x)) + C
1/6 (x + 1/2 (sin x $$\sqrt{1 - sin ^{2}x}$$)) + C
1/6 (x + 1/2 (3t$$\sqrt{1 - 9t^{2}}$$)) + C

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

2. Mar 5, 2009

### yyat

I think you will have to use the arcsine function, which is the inverse of sine.
Also, there is a small error in your calculations: sin(2x) is not sin(x)cos(x).

3. Mar 5, 2009

### djeitnstine

t = 1/3 sin x

$$x= sin^{-1}3t$$

its just algebra.

Yes $$sin2x = 2sin(x)cos(x)$$ so the 1/2 goes away

4. Mar 5, 2009

### mvantuyl

This is weird.

The original statement of the problem was:

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 replied:

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

The Bob

Perfect! Thanks.

All that is now gone.