Question

Integration of log sin2x from 0 to π/2

Answer 1

$I = \int\limits^{\pi/2}_0 {log(sin2x) \, dx$

We know, sin2x = 2sinx.cosx
So, log(sin2x) = log(2sinx.cosx) = log2 + logsinx + logcosx
so, $I = \int\limits^{\pi/2}_0 {log(sin2x) \, dx =\int\limits^{\pi/2}_0{log2}\,dx+\int\limits^{\pi/2}_0{log(sinx)\,dx+\int\limits^{\pi/2}_0{log(cosx)\,dx$
Now, we have to find I$I = \int\limits^{\pi/2}_0 {log(sinx) \, dx$ and $I = \int\limits^{\pi/2}_0 {log(cosx) \, dx$

$I = \int\limits^{\pi/2}_0 {log(sinx) \, dx =\int\limits^{\pi/2}_0{log(sin(\pi/2-x)}\,dx$
$I = \int\limits^{\pi/2}_0 {log(cosx) \, dx$
$2I = \int\limits^{\pi/2}_0 {log(cosx)+log(sinx)}\, dx =\int\limits^{\pi/2}_0{log(sin2x/2)}\,dx$
$=\int\limits^{\pi/2}_0[{log(sin2x)-log2}]\,dx=\int\limits^{\pi/2}_0{logsin2x}\,dx-\int\limits^{\pi/2}_0{log2}\,dx$
$=1/2\int\limits^{\pi}_0{log(sinx)\,dx-\int\limits^{\pi/2}_0{log2}\,dx$
$=2/2\int\limits^{\pi/2}_0{log(sinx)\,dx-\int\limits^{\pi/2}_0{log2}\,dx$
2I = I $-\pi/2log2$
I = -π/2log2

Similarly , I = $\int\limits^{\pi/2}_0{log(cosx)\,dx = -\pi/2log2$$\int\limits^{\pi/2}_0{log(cosx)\,dx = -\pi/2log2$

Then, $\int\limits^{\pi/2}_0{log(sin2x)\,dx = -\pi/2log2 - \pi/2log+\pi/2log2=\pi/2log2$

Hence, answer is -π/2 log2