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 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 \int1 + 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.

 

[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).

 

[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

 

[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\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 replied:

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

The Bob

And I answered:

Perfect! Thanks.

All that is now gone.