# Intergration using trig substitution

• Nov 14th 2010, 05:37 PM
Frostking
Intergration using trig substitution
I need to integrate 1/ (1 - x^2)^(1/2) dx

I know I can use x = sin theta but... I was also taught that I can use x = cos theta

When I do this I end up with the integral of - sin theta/ sin theta therefore
the antiderv is negative theta when I know it should be theta!!!! I know I then need to put the answer back in terms of x and add C. Using the sin sub I get arcsin of x + When I use cos sub I get - arccos of x + C!!!!!Could someone please tell me where I am making my mistake? Thanks again for all the time you folks devote! Frostking
• Nov 14th 2010, 05:53 PM
Quote:

Originally Posted by Frostking
I need to integrate 1/ (1 - x^2)^(1/2) dx

I know I can use x = sin theta but... I was also taught that I can use x = cos theta

When I do this I end up with the integral of - sin theta/ sin theta therefore
the antiderv is negative theta when I know it should be theta!!!! I know I then need to put the answer back in terms of x and add C. Using the sin sub I get arcsin of x + When I use cos sub I get - arccos of x + C!!!!!Could someone please tell me where I am making my mistake? Thanks again for all the time you folks devote! Frostking

You are not making any mistakes. They are both correct. If we introduce limits into this integral,

$\displaystyle \int^{1}_{-1}\frac{1}{\sqrt{1-x^2}}$

Solution(1), arcsin 1- arcsin(-1) = 180 degrees.

Solution(2), -arccos 1 + arccos (-1) =180 degrees.

They both arrive at the same result.
• Nov 14th 2010, 05:55 PM
skeeter
Quote:

Originally Posted by Frostking
I need to integrate 1/ (1 - x^2)^(1/2) dx

I know I can use x = sin theta but... I was also taught that I can use x = cos theta

When I do this I end up with the integral of - sin theta/ sin theta therefore
the antiderv is negative theta when I know it should be theta!!!! I know I then need to put the answer back in terms of x and add C. Using the sin sub I get arcsin of x + When I use cos sub I get - arccos of x + C!!!!!Could someone please tell me where I am making my mistake? Thanks again for all the time you folks devote! Frostking

note the following derivatives ...

$\displaystyle \displaystyle \frac{d}{dx} \arcsin{x} = \frac{1}{\sqrt{1-x^2}}$

$\displaystyle \displaystyle \frac{d}{dx} \arccos{x} = -\frac{1}{\sqrt{1-x^2}}$

what's that tell you?