I am interested in knowing different ways to find reduction formulae for the following integral .
$I_{n}= \int_{0}^{\frac{\pi}{2}} \cos 2nx\cdot ln\cos x$
What I did was make a substitution $ln\cos x=t$ and used IBP , so does anyone have an easier or perhaps a more interesting method to find the reduction formulae ?
And yes , btw the way the reduction formula comes out to be $ I_{n}= (-1)^{n-1} \frac{\pi}{4n}$
You can find many answers on MSE where the Fourier series of $\log\cos x$ is mentioned.
Over $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ we have:
$$ \log\cos x = -\log 2-\sum_{n\geq 1}\frac{(-1)^n \cos(2nx)}{n} $$ hence: $$ I_n = \frac{\pi}{4n}(-1)^{n+1}.$$ Anyway, integration by parts and induction are just enough. Consider that $\frac{d}{dx}\log\cos x=-\tan x$.