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Integration of log(sin2x ) limit 0 to pi/4?
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Integration of log(sin2x ) limit 0 to pi/4?
Integration of log(sin2x) limit 0 to pi/4


When solving integrals, it is important to first identify the function to be integrated. In this case, we are integrating the function log(sin2x) from 0 to pi/4.


Step 1: Simplify the function


We can simplify the function by using the identity sin2x = 2sinx cosx. Therefore, we have:


log(sin2x) = log(2sinx cosx) = log2 + log(sinx) + log(cosx)


Step 2: Integrate each term separately


Now that we have simplified the function, we can integrate each term separately. Using the formula for the integral of log(x), we have:



  • ∫log2 dx = x log2 - x + C

  • ∫log(sinx) dx = -x log2 - x + 2x log(sin(pi/4)) + C = -x log2 - x + 2x log(1/sqrt(2)) + C

  • ∫log(cosx) dx = -x log2 - x + 2x log(cos(pi/4)) + C = -x log2 - x + 2x log(1/sqrt(2)) + C



where C is the constant of integration.


Step 3: Evaluate the definite integral


Now that we have found the indefinite integral of each term, we can evaluate the definite integral from 0 to pi/4. Plugging in the limits of integration, we get:



  • ∫log2 dx (from 0 to pi/4) = (pi/4) log2 - (pi/4)

  • ∫log(sinx) dx (from 0 to pi/4) = -(pi/4) log2 - (pi/4) + pi/2 log(1/sqrt(2))

  • ∫log(cosx) dx (from 0 to pi/4) = -(pi/4) log2 - (pi/4) + pi/2 log(1/sqrt(2))



Step 4: Combine the terms


Finally, we can combine the terms to get the value of the definite integral:



  • ∫log(sin2x) dx (from 0 to pi/4) = (pi/2) log(1/sqrt(2)) - pi/2



Therefore, the value of the definite integral is (pi/2) log(1/sqrt(2)) - pi/2.
Community Answer
Integration of log(sin2x ) limit 0 to pi/4?
Integration log(sin2x) =-cos2x/2 putting the limits, the answer is 1/2
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Integration of log(sin2x ) limit 0 to pi/4?
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