# Math Help - integral expression for bessel functions...

1. ## integral expression for bessel functions...

Can someone give me a hand with this problem? I cant figure out how to integrate things once I have y y' and y'' plugged into the bessel function.

problem in a .png linked below

I only need help with the integration on part A. Thanks in advance!

-Canopy

2. Originally Posted by canopy
Can someone give me a hand with this problem? I cant figure out how to integrate things once I have y y' and y'' plugged into the bessel function.

problem in a .png linked below

I only need help with the integration on part A. Thanks in advance!

-Canopy
$y = \int_0^{\pi/2} \cos( x \sin \theta) d \theta$
$y' = \int_0^{\pi/2} - \sin \theta \sin( x \sin \theta) d \theta$
$y'' = \int_0^{\pi/2} - \sin^2 \theta \cos( x \sin \theta) d \theta$

so

$x^2y'' + x y' + x^2 y =$ $\int_0^{\pi/2} - x^2\sin^2 \theta \cos( x \sin \theta) - x \sin \theta \sin( x \sin \theta) + x^2 \cos( x \sin \theta) d \theta$
$=\int_0^{\pi/2} x^2(1-\sin^2) \theta \cos( x \sin \theta) - x \sin \theta \sin( x \sin \theta) d \theta$
$=\int_0^{\pi/2} x^2\cos^2 \theta \cos( x \sin \theta) - x \sin \theta \sin( x \sin \theta) d \theta$
$= \int_0^{\pi/2} d\left( x \cos \theta \sin( x \sin \theta) \right) = \left. x \cos \theta \sin( x \sin \theta) \right|_0^{\pi/2} = 0$