Recursion in Integration by Parts

Question

I'm trying to integrate by parts, but I keep getting recursive answers.

$I=\int_0^\pi f(x)cos(x)dx$ where $f''(x)=3f(x)$, $f'(0)=-5$, and $f'(\pi)=4$

Thanks.

Answer 1

The recursion will eventually be in terms of itself, which means you can solve for $I$. So: \begin{align*} I = \int_0^\pi f(x)\cos x\,dx &= f(x) \sin x\, \bigg|_0^\pi - \int_0^\pi f'(x)\sin x\,dx \\ &= 0 - \bigg( f'(x)(-\cos x)\, \bigg|_0^\pi - \int_0^\pi f''(x)(-\cos x)\,dx \bigg) \\ &= f'(\pi)\cos \pi - f'(0)\cos 0 - \int_0^\pi 3f(x)\cos x\,dx \bigg) \\ &= (-5)(-1) - 4(0) - 3I. \end{align*} Therefore $4I=1$, or $I=\frac14$.