Question

Log(sin2x)dx limit 0 to pi/4

Answer 1

https://www.quora.com/What-is-the-integration-of-0-to-pi-2-log-sin2x

bro here you will get your answer.

Answer 2

$\boxed{\int_0^{\pi/4} \log(\sin 2x)dx=-\frac{\pi}{4}\log 2}$

Step-by-step explanation:

$I=\int_0^{\pi/4} \log(\sin 2x)dx$

$=\int_0^{\pi/4}\log(\sin2(\frac{\pi}{4}-x))dx$  (∵ $\int_0^a f(x)dx=\int_0^a f(a-x)dx$)

$=\int_0^{\pi/4}\log(\sin(\frac{\pi}{2}-2x))dx$

$=\int_0^{\pi/4}\log(\cos 2x)dx$

Therefore,

$I+I=\int_0^{\pi/4} \log(\sin 2x)dx+\int_0^{\pi/4}\log(\cos 2x)dx$

$\implies 2I=\int_0^{\pi/4}[\log(\cos 2x)+\log(\sin 2x)]dx$

$\implies 2I=\int_0^{\pi/4}[\log(\cos 2x\times \sin 2x)]dx$  (by log property)

$\implies 2I=\int_0^{\pi/4}[\log(\frac{2}{2}\times\cos 2x\times \sin 2x)]dx$

$\implies 2I=\int_0^{\pi/4}[\log(2\times\cos 2x\times \sin 2x)-\log 2]dx$

$\implies 2I=\int_0^{\pi/4}[\log(\sin 4x)-\log 2]dx$

$\implies 2I=\int_0^{\pi/4}\log(\sin 4x)dx-\int_0^{\pi/4}\log 2dx$

taking 2x = t

2 dx = dt

When $x=0, t=0$

When $x=\pi/4, t=\pi/2$

Thus,

$2I=\frac{1}{2}\int_0^{\pi/2}\log(\sin 2t)dt-\log 2 \Bigr|x\Bigr|_0^{\pi/4}$

$\implies 2I=2\times\frac{1}{2}\int_0^{\pi/4}\log(\sin 2x)dx-\frac{\pi}{4}\log 2$

$\implies 2I=\int_0^{\pi/4}\log(\sin 2x)dx-\frac{\pi}{4}\log 2$

$\implies 2I=I-\frac{\pi}{4}\log 2$

$\implies I=-\frac{\pi}{4}\log 2$

Hope this answer is helpful.

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