# Function given by an integral satisfies Bessel differential equation

Show that $f(x)=\int_{0}^{\pi}\cos{(x\cos{\theta})}\sin^{2n}{\theta} d \theta$, where $n \in \mathbb{N}_0$, satisfies Bessel differential equation

$x^2{f}''+x{f}'+(x^2-n^2)f=0$

I wanted to calculate both derivatives of the function and then show that they satisfy the equation, but how do I calculate the derivative of a function given by an integral?

• i think you should first review your calculus lectures before you start with something as complicated as bessel functions...:) – tired Oct 13 '16 at 12:49

$$\frac{d}{dx} \int_a^b f(x,y) dy = \int_a^b \frac{\partial f}{\partial x}(x,y) dy.$$