integral log [sin x]
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Answered by
37
The indefinite Integral of Log (Sin x) dx has no simple function form. That can only be expressed in terms of complex functions.
It can be found by Taylor series expansion or Mclaurin series.
But definite integral from 0 to pi/2 can be found to be - pi/2 * Ln 2..

If we want to evaluate the definite integral then:
![I=\int \limits_0^{\pi/2} {LogSinx}\ dx\\\\=\int \limits_{\pi/2}^0 {LogSin(\frac{\pi}{2}-x})\ d(\frac{\pi}{2}-x)\\\\So\ I=\int \limits_0^{\pi/2} {LogCosx}\ dx\\\\I+I=\int \limits_0^{\pi/2} {Log\frac{Sin2x}{2}}\ dx\\\\=\frac{1}{2} \int \limits_{2x=0}^{\pi}LogSin2x\ d(2x)-\int \limits_0^{\pi/2}Log2\ dx\\\\=\frac{1}{2} [\int \limits_0^{\pi/2} {LogSinx}\ dx+\int \limits_{\pi/2}^{\pi} {LogSinx}\ dx]-\pi/2*Log2 I=\int \limits_0^{\pi/2} {LogSinx}\ dx\\\\=\int \limits_{\pi/2}^0 {LogSin(\frac{\pi}{2}-x})\ d(\frac{\pi}{2}-x)\\\\So\ I=\int \limits_0^{\pi/2} {LogCosx}\ dx\\\\I+I=\int \limits_0^{\pi/2} {Log\frac{Sin2x}{2}}\ dx\\\\=\frac{1}{2} \int \limits_{2x=0}^{\pi}LogSin2x\ d(2x)-\int \limits_0^{\pi/2}Log2\ dx\\\\=\frac{1}{2} [\int \limits_0^{\pi/2} {LogSinx}\ dx+\int \limits_{\pi/2}^{\pi} {LogSinx}\ dx]-\pi/2*Log2](https://tex.z-dn.net/?f=I%3D%5Cint+%5Climits_0%5E%7B%5Cpi%2F2%7D+%7BLogSinx%7D%5C+dx%5C%5C%5C%5C%3D%5Cint+%5Climits_%7B%5Cpi%2F2%7D%5E0+%7BLogSin%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%7D%29%5C+d%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%29%5C%5C%5C%5CSo%5C+I%3D%5Cint+%5Climits_0%5E%7B%5Cpi%2F2%7D+%7BLogCosx%7D%5C+dx%5C%5C%5C%5CI%2BI%3D%5Cint+%5Climits_0%5E%7B%5Cpi%2F2%7D+%7BLog%5Cfrac%7BSin2x%7D%7B2%7D%7D%5C+dx%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D+%5Cint+%5Climits_%7B2x%3D0%7D%5E%7B%5Cpi%7DLogSin2x%5C+d%282x%29-%5Cint+%5Climits_0%5E%7B%5Cpi%2F2%7DLog2%5C+dx%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D+%5B%5Cint+%5Climits_0%5E%7B%5Cpi%2F2%7D+%7BLogSinx%7D%5C+dx%2B%5Cint+%5Climits_%7B%5Cpi%2F2%7D%5E%7B%5Cpi%7D+%7BLogSinx%7D%5C+dx%5D-%5Cpi%2F2%2ALog2)
![2I=\frac{1}{2}[I+I]-\pi/2*Log2\\\\I=-\frac{\pi}{2}Log2 2I=\frac{1}{2}[I+I]-\pi/2*Log2\\\\I=-\frac{\pi}{2}Log2](https://tex.z-dn.net/?f=2I%3D%5Cfrac%7B1%7D%7B2%7D%5BI%2BI%5D-%5Cpi%2F2%2ALog2%5C%5C%5C%5CI%3D-%5Cfrac%7B%5Cpi%7D%7B2%7DLog2)
It can be found by Taylor series expansion or Mclaurin series.
But definite integral from 0 to pi/2 can be found to be - pi/2 * Ln 2..
If we want to evaluate the definite integral then:
Answered by
14
Answer:
Tan x
Step-by-step explanation:
Log sinx
=(1/sinx) *cosx
=Tanx
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