$\int_{\theta}^{\theta+\pi}r(u)\cos u\, du=-\lambda\sin\theta$ and $\int_\theta^{\theta+\pi}r(u)\sin u\;du=\lambda\cos\theta$ for certain $r(u)$

Question

We are given some definitions about a function $r(u)$:

1) $r(u)>0$

2) $r(u)= r(u+2\pi)$

3) $r(u)+r(u+\pi)= \lambda$

4) $r(u)$ is continuous

5) $\int_0^{\pi}r(u)\cos\theta=0$ and $\int_0^{\pi}r(u)\sin\theta=\lambda$

The following identities are to be proven:

The first proceeds as follows:

I have tried over and over to prove the second identity, and it seems like it should be easy. However, I keep getting $\lambda +\lambda\cos\theta$ instead of $\lambda\cos\theta$ ... what am I missing here? And sorry for the awful formatting, as you can see I'm fairly new to MSE.

Answer 1

Let $$F(\theta)=\int_{\theta}^{\theta+\pi}r(u)\sin(u)du.$$

The derivative is

$$F'(\theta)=$$

$$r(\theta+\pi)\sin(\theta+\pi)-r(\theta)\sin(\theta)=$$

$$-\sin(\theta)\Bigl(r(\theta+\pi)+r(\theta)\Bigr)=$$

$$-\lambda \sin(\theta)$$

and after integration,

$$F(\theta)=\lambda \cos(\theta)+C$$

with $\theta=0$, it becomes $$\int_0^\pi r(u)\sin(u)du=\lambda=\lambda+C$$ thus $$C=0$$ Done.