Free help with homework

Why join Brainly?

  • ask questions about your assignment
  • get answers with explanations
  • find similar questions



This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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..

Repetitively\ Apply\ \int{u}\ dv=uv-\int {v}\ du\\\\\int Log\ sinx\ dx=x\ LogSinx-\int {x\ cotx}\ dx\\\\=x\ logSinx-x^2/2\ Cotx+1/2 \int\ x^2\ cosec^2x\ dx\\\\=x\ logSinx-x^2/2\ Cotx+x^3/6\ cosec^2x+1/3\ \int {x^3 cosec^2x\ cotx}\ dx\\\\x\ logSinx-x^2/2\ Cotx+x^3/6\ cosec^2x+x^4/12\ cosec^2x\ cotx+.....

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

1 5 1
The Brain
  • The Brain
  • Helper
Not sure about the answer?
Learn more with Brainly!
Having trouble with your homework?
Get free help!
  • 80% of questions are answered in under 10 minutes
  • Answers come with explanations, so that you can learn
  • Answer quality is ensured by our experts