Substitution rule uv-∫vdu?

Question

Given that $\int_0^{\pi}(f(x) + f''(x)) \sin x dx = 2$ and $f(\pi) = 1$ Find $f (0)$. Do we can solve with $uv-\int \!vdu$? How? I couldn't find a good result.

Answer 1

Hints.

$$\int\left(f(x) + f''(x)\right) \sin(x)\ dx = \int f(x)\sin(x)\ dx + \int f''(x)\sin(x)\ dx$$

$$\int f(x)\sin(x)\ dx = \int u(x) v'(x) = u(x) v(x) - \int u'(x) v(x)\ dx$$

With $u = f(x)$, $v = \sin(x)$.

Try to play a bit.

Answer 2

Break the integral: $$I=\int_0^{\pi} f(x)\sin xdx+\int_0^{\pi} f''(x)\sin xdx=A+B=2.$$ Integrate $A$ by parts twice: $$A=-f(x)\cos x|_0^{\pi}\color{red}{+}\int_0^{\pi} f'(x)\cos xdx=f(\pi)+f(0)\color{red}{+}f'(x)\sin x|_0^{\pi}-\int_0^{\pi} f''(x)\sin xdx=1+f(0)\color{red}{+}0-B.$$ Now: $$A+B=1+f(0)-B+B=2 \Rightarrow f(0)=1.$$