# Integral involving trigonometric functions

I am learning a theory involving the Bessel functions and I encountered an integral equality that I could not prove.

Let $$z$$ be a fixed complex number and $$n$$ an odd number.

The author wrote that, $$\frac{1}{\pi} \int_0^\pi \sin n\theta \sin (z\sin \theta)\,d\theta=\frac{2}{\pi} \int_0^{\pi/2} \sin n\theta \sin (z\sin \theta)\,d\theta .$$

Therefore I wanted to prove that $$\int_0^{\pi/2} \sin n\theta \sin (z\sin \theta)\,d\theta=\int_{\pi/2}^{\pi} \sin n\theta \sin (z\sin \theta)\,d\theta.$$

But I failed doing so.

My attempt:

By change of variables, $$\theta \mapsto \theta + \pi/2,$$ we have

$$\int_{\pi/2}^{\pi} \sin n\theta \sin (z\sin \theta)\,d\theta=\int_{0}^{\pi/2} \sin (n\theta+\frac{n\pi}{2}) \sin (z\sin (\theta+\frac{\pi}{2}))\,d\theta= \int_{0}^{\pi/2} \cos n\theta \sin\frac{n\pi}{2} \sin (z\cos \theta)\,d\theta.$$

Note that $$\sin\frac{n\pi}{2}$$ can be either $$1$$ or $$-1$$ and it depends on an integer $$m$$ which depends on $$n$$, where $$m\equiv n$$ (mod $$4$$).

However, the result is still much different from the desired equality that I wanted. Did I miss something simple or what?

Any help will be appreciated.

Well, I managed a way to prove the desired integral equality. The proof is all about elementary but tedious trigonometric identities.

Let $$n$$ be an odd integer.

A direct calculation with appealing to elementary trigonometric identities gives $$\sin (n\theta +\frac{n\pi}{2})=-\sin (n\theta - \frac{n\pi}{2}).$$

Therefore by the equality above and the change of variables, $$\theta \mapsto \theta+\frac{\pi}{2},$$ we have

\begin{align*} & \int_0^{\pi/2 } \sin (n\theta +\frac{n\pi}{2}) \sin(z\sin(\theta+\frac{\pi}{2}))d\theta\\ & =- \int_0^{\pi/2 } \sin (n\theta -\frac{n\pi}{2}) \sin(z\sin(\theta+\frac{\pi}{2}))d\theta\\ & =- \int_{-\pi/2 }^0 \sin n\theta\sin(z\sin(\theta+\pi))d\theta\\ & = \int_{-\pi/2 }^0 \sin n\theta\sin(z\sin\theta)d\theta. \end{align*}

Applying change of variables $$\theta \mapsto -\theta$$, then we derive $$\int_{-\pi/2 }^0 \sin n\theta\sin(z\sin\theta)d\theta= \int_0^{\pi/2 } \sin n\theta\sin(z\sin\theta)d\theta.$$

Hence, we have proved $$\int_0^{\pi/2 } \sin (n\theta +\frac{n\pi}{2}) \sin(z\sin(\theta+\frac{\pi}{2}))d\theta= \int_0^{\pi/2 } \sin n\theta\sin(z\sin\theta)d\theta,$$ where is the desired integral equality.