Is $\int^{\pi}_0|e^{iRcos\theta-Rsin\theta}|d\theta=\int_0^{\pi}e^{-Rsin\theta}d\theta$?

Question

I'm looking over my notes and I noticed a few dubious lines of reasoning ( I will take it to my lecturer as well for further clarification but I won't be able to do that for a few days )

Specifically it's an example for solving real integrals using complex analysis.The integral is :

$$\int^{\infty}_0 \tfrac{sin(x)}{x}dx$$

There is a part which states

$$\int^{\pi}_0|e^{iRcos\theta-Rsin\theta}|d\theta=\int_0^{\pi}e^{-Rsin\theta}d\theta$$

why is this ? the square root of the square of integrand doesn't equal the left hand side integrand according to my calculations.....Maybe I'm making a stupid mistake

Answer 1

$$|e^{iR\cos \theta-R\sin \theta}|=|e^{iR\cos \theta}e^{-R\sin \theta}|=|e^{iR\cos \theta}||e^{-R\sin \theta}|=1\times|e^{-R\sin \theta}|$$

Answer 2

What can we say about $|e^{iR\cos(\theta)}|$? Note that the real part of $e^{iR\cos(\theta)}$ squared plus the imaginary part squared must be 1. (Why is this?). Hint: It has to do with your comment on the top answer.