Let f be continuous on [a,b]. Show that:
\displaystyle \int_a^b f(x) \mathrm dx = \int_a^b f(a + b - x) \mathrm dx
Rationalise this geometrically.
(i) Find:
\displaystyle \int_0^{\pi/2} \frac {\sin^{2020} x} {\cos^{2020} x + \sin^{2020} x} \mathrm dx
(ii) Find:
\displaystyle \int_0^{\pi/2} \log(1 + \tan x) \mathrm dx
(iii) Find:
\displaystyle \int_0^{\pi/2} \log \sin x \mathrm dx
[all classic exercises but (ii) and (iii) were posted on the hard integral thread by DFranklin]
Consider the substitution y=a+b-x then dy=-dx
\displaystyle\int^b_a f(y) dy= \int^a_b -f(a+b-x) dx= \int^b_a f(a+b-x) dx
\int^b_a f(x) dx is the area under the curve between a and b starting from the left to the right while \int^b_a f(a+b-x) dx is the area under the curve when integrating form the right to the left.
i) Using \sin(\frac\pi2-x)=\cos x, \cos(\frac\pi2-x)=\sin x, and the stem we have:
\displaystyle \int_0^{\pi/2} \frac {\sin^{2020} x} {\cos^{2020} x + \sin^{2020} x} \mathrm dx= \int_0^{\pi/2} \frac {\cos^{2020} x} {\cos^{2020} x + \sin^{2020} x} \mathrm dx
Letting I=\displaystyle \int_0^{\pi/2} \frac {\sin^{2020} x} {\cos^{2020} x + \sin^{2020} x} \mathrm dx thus:\
2I=\displaystyle \int_0^{\pi/2} \frac {\cos^{2020} x + \sin^{2020} x} {\cos^{2020} x + \sin^{2020} x} \mathrm dx=\int_0^{\pi/2}1 \ \mathrm dx= {\pi \over 2}
\displaystyle \therefore \ I=\frac\pi4
ii) \displaystyle \int_0^{\pi/4} \log(1 + \tan x) \mathrm dx=\displaystyle \int_0^{\pi/4} \log\left(1 + \tan\left(\frac\pi4- x\right)\right) \mathrm dx
\displaystyle 1+\tan\left(\frac\pi4- x\right)=1+{1-\tan x \over 1+\tan x}
\displaystyle={2 \over 1+\tan x}
\displaystyle \log\left({2 \over 1+\tan x}\right)=\log 2 \ - \log(1+\tan x)
\therefore \displaystyle \int_0^{\pi/4} \log\left(1 + \tan x\right) \mathrm dx =\displaystyle \int_0^{\pi/4} \log\left(1 + \tan\left(\frac\pi4- x\right)\right) \mathrm dx= \frac\pi4\log 2 -\displaystyle \int_0^{\pi/4} \log\left(1 + \tan x\right) \mathrm dx
\displaystyle \therefore \ 2\int_0^{\pi/4} \log(1 + \tan x) \mathrm dx=\frac\pi4\log 2
\displaystyle \int_0^{\pi/4} \log(1 + \tan x) \mathrm dx=\frac\pi8\log 2
iii) Let K=\displaystyle \int_0^{\pi/2} \log \sin x \mathrm dx
K=\displaystyle \int_0^{\pi/2} \log \cos x \mathrm dx
2K=\displaystyle \int_0^{\pi/2} \log \cos x +\log \sin x\mathrm dx
=\displaystyle \int_0^{\pi/2} \log (\sin x\cos x) \ \mathrm dx
=\displaystyle \int_0^{\pi/2} \log \sin 2x +\log {1 \over 2} \mathrm dx\\
=\int_0^{\pi/2} \log \sin 2x \ \mathrm dx-\frac\pi2\log2
\displaystyle =\int_0^{\pi/4} \log \sin 2x \ \mathrm dx+\int_{\pi/4}^{\pi/2} \log \sin 2x \ \mathrm dx-\frac\pi2\log2
Letting y=2x we have dx=\frac12dy
Thus, \frac12\int_0^{\pi/2} \log \sin y \ \mathrm dy=\int_0^{\pi/4} \log \sin 2x \ \mathrm dx
Letting z=2x-\frac\pi2 we have dx=\frac12dz
Therefore, \frac12\int_0^{\pi/2} \log \sin \left(z+\frac\pi2\right) \ \mathrm dz=\int_{\pi/4}^{\pi/2} \log \sin 2x \ \mathrm dx
Since \sin \left(x+\frac\pi2\right)=\cos x we have:
\displaystyle 2K=\frac12\int_0^{\pi/2} \log \sin x \ \mathrm dx \ +\frac12\int_0^{\pi/2} \log \cos x \mathrm dx-\frac\pi2\log2
\displaystyle 2K=K-\frac\pi2\log2
K=-\frac\pi2\log2
oops, yes I did mean \dfrac \pi 4 in that upper bound, sorry! \dfrac \pi 2 still has a closed form in terms of Catalan’s constant.
One that’s more in the flavour of this thread would be \displaystyle \int_0^{\pi/2} \ln \tan \theta \mathrm d\theta.
Can’t edit the OP though.
Okay, added that one.
Also I found \displaystyle \int^{\pi/2}_0 \ln\tan x \ \mathrm dx unintentionally like this:
\tan(\frac\pi2-x)=\cot x
\displaystyle \int_0^{\pi/2} \log(1 + \tan x) \mathrm dx=\displaystyle \int_0^{\pi/2} \log(1 + \cot x) \mathrm dx
\displaystyle =\int_0^{\pi/2} \log\left({\tan x +1 \over \tan x}\right) \mathrm dx
\displaystyle =\int_0^{\pi/2} \log(\tan x +1) - \log\tan x \ \mathrm dx
Therefore,
\displaystyle \int_0^{\pi/2} \log(1 + \tan x) \mathrm dx=\int_0^{\pi/2} \log(\tan x +1) \ \mathrm dx - \int^{\pi/2}_0\log\tan x \ \mathrm dx
\therefore \displaystyle \int^{\pi/2}_0\log\tan x \ \mathrm dx=0
Alternatively since \displaystyle \tan \left(\frac \pi 2 - x\right) = \cot x you have \displaystyle 2 \int_0^{\pi/2} \log \tan x \mathrm dx = \int_0^{\pi/2} \log (\tan x \cot x) \mathrm dx = 0
good work