$\int^{2\pi}_{0}g(x)\sin(x)dx +\int^{2\pi}_{0}g''(x)\sin(x)dx = 2$. Given that $g(2\pi) = 1$, prove that $g(0)=3$

Question

Suppose that $g''(x)$ is continuous everywhere and that

$$\int^{2\pi}_{0}g(x)\sin(x)dx +\int^{2\pi}_{0}g''(x)\sin(x)dx = 2$$

Given that $g(2\pi) = 1$, prove that $g(0)=3$

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I tried to integrate by parts, the first integral.

Let $u = g(x)$

$du = g'(x) dx$

Let $dv=\sin(x)dx$

$v=-\cos(x)$

$$=uv - \int^{2\pi}_{0}vdu$$ $$=-g(x)\cos(x) + \int^{2\pi}_{0}g'(x)\cos(x)dx$$

Now I can apply parts again.

Let $u = g'(x)$

$du = g''(x) dx$

Let $dv=\cos(x)dx$

$v=\sin(x)$

$$=uv - \int^{2\pi}_{0}vdu$$ $$=g'(x)\sin(x) -\int^{2\pi}_{0}g''(x)\sin(x)dx$$

Adding these two together to get the first integral, I get:

$$-g(x)\cos(x) + (g'(x)\sin(x) -\int^{2\pi}_{0}g''(x)\sin(x)dx)$$

$$-g(x)\cos(x) + g'(x)\sin(x) -\int^{2\pi}_{0}g''(x)\sin(x)dx$$

Adding to the second integral, I get:

$$-g(x)\cos(x) + g'(x)\sin(x) -\int^{2\pi}_{0}g''(x)\sin(x)dx + \int^{2\pi}_{0}g''(x)\sin(x)dx$$

$$-g(x)\cos(x) + g'(x)\sin(x) = 2$$

After this I don't know how to proceed. Any suggestions?

Answer 1

Hint: let $h(x) = g'(x) \sin x - g(x) \cos x$. Then desired integral is $h(2\pi)-h(0)=2$.

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Another approach more in line with your efforts is to use integration by parts (twice) and show that $$\int_{0}^{2\pi}g''(x)\sin x\, dx=-\int_{0}^{2\pi}g'(x)\cos x\, dx=g(0)-g(2\pi)-\int_{0}^{2\pi}g(x)\sin x\, dx$$ so that $g(0)-g(2\pi)=2$.