1. ## Integration Help

hi, im havin some problems solving this integral,

if $y= \int_0^\frac{1}{2\pi} \cos(x \sin \theta) d\theta$

show that $xy'' + y' +xy =0$

ive found the integral to be $\frac{\sin(x \sin \theta)} {-x \cos\theta}$ is this correct, i found it using the substitution method where $u = x \sin\theta$

if this integral is correct, how do i solve the rest of the equation?

2. Originally Posted by iLikeMaths
hi, im havin some problems solving this integral,

if $y= \int_0^\frac{1}{2\pi} \cos(x \sin \theta) d\theta$

show that $xy'' + y' +xy =0$

ive found the integral to be $\frac{\sin(x \sin \theta)} {-x \cos\theta}$ is this correct, i found it using the substitution method where $u = x \sin\theta$

if this integral is correct, how do i solve the rest of the equation?
$y= \int_0^\frac{1}{2\pi} \cos(x \sin \theta) d\theta$

Use parts first.

$y= \int_0^\frac{1}{2\pi} \cos(x \sin \theta) d\theta = \bigg(\theta \cos(x \sin \theta)\bigg)\bigg|_0^\frac{1}{2\pi} - \int \theta (-\sin(x\sin\theta)x\cos(\theta))d\theta$

$= \bigg(\theta \cos(x \sin \theta)\bigg)\bigg|_0^\frac{1}{2\pi} + \int \theta \sin(x\sin\theta)x\cos(\theta)d\theta$

Now use a substitution for the intergral.

$u = x\sin(\theta)$

$\frac{du}{d\theta} = x\cos(\theta)$

$du = x\cos(\theta)d\theta$

$= \bigg(\theta \cos(x \sin \theta)\bigg)\bigg|_0^\frac{1}{2\pi} + \int \theta \sin(u)du$

And from $u = x\sin(\theta)$, we can get $\theta = \asin\big\frac{u}{x}zbig)$

Integral is:

$= \bigg(\theta \cos(x \sin \theta)\bigg)\bigg|_0^\frac{1}{2\pi} + \int \arcsin \big(\frac{u}{x}\big) \sin(u)du$

3. when u integrated by parts, what did u use as ur "u" and "v" values because there seems to be a random $\theta$ and i dont know where it came from, please explain, and also is that the complete integral or do i still have to integrate the part where there is $\arcsin$thanks for ur post