# Finding I(x)

• Feb 14th 2009, 12:38 PM
nmatthies1
Finding I(x)
Okay so the question goes

For x in [-pi/2,pi/2], define $I(x)=\int_{sin(x)}^{cos(x)}\frac {1} {t+(\sqrt(1-t^2))} dt$

Find I(x) by showing that dI/dx = -1 and that I(pi/4)=0.

I have the done the first part, ie showing that dI/dx= -1 and that I(pi/4)=0. However, I can't figure out how to use my answer to that part to find I(x). Can anyone help?
• Feb 14th 2009, 12:52 PM
skeeter
Quote:

Originally Posted by nmatthies1
Okay so the question goes

For x in [-pi/2,pi/2], define $I(x)=\int_{sin(x)}^{cos(x)}\frac {1} {t+(\sqrt(1-t^2))} dt$

Find I(x) by showing that dI/dx = -1 and that I(pi/4)=0.

I have the done the first part, ie showing that dI/dx= -1 and that I(pi/4)=0. However, I can't figure out how to use my answer to that part to find I(x). Can anyone help?

looks like you did all the hard work ...

$\frac{dI}{dx} = -1$

$I(x) = -x + C$

$I \left(\frac{\pi}{4}\right) = 0$

$I(x) = \frac{\pi}{4} - x$
• Feb 14th 2009, 12:54 PM
nmatthies1
Oh yes of course! I was looking for something harder I guess. Thanks!