posted by .

Integral of ln(sinx+cosx) with respect to x from -pi/4 to pi/4

  • calculus -

    sin(x) + cos(x) = sqrt(2) sin(x + pi/4)

    So, the integral can be written as:

    pi/4 ln(2) + Integral of Log[sin(x)]dx from 0 to pi/2

    Let's call the integral in here I:

    I = Integral of Log[sin(x)]dx from 0 to pi/2

    Then note that because sin is symmetric w.r.t. reflection about x = pi/2 integrating to pi will yield twice the value:

    2 I = Integral of Log[sin(x)]dx from 0 to pi

    Then substitute in this integral
    x = 2 y:

    2 I = 2 Integral of Log[sin(2y)]dy from 0 to pi/2 ----->

    I = Integral of
    (Log(2) + Log[cos(y)] +Log[sin(y)] )dy from 0 to pi/2 =

    pi/2 Log(2) + 2 I

    because the integral of log[cos(y)] is the same as the integral of
    log[sin(y)], as cos(y) is sin(pi/2-y).

    Solving for I gives:

    I = -pi/2 log(2)

    The original integral is thus
    - pi/4 log(2)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Math

    Verify the identity . (cscX-cotX)^2=1-cosX/1+cosX _______ sorry i cant help you (cscX-cotX)=1/sinX - cosX/sinX = (1-cosX)/sinX If you square this you have (1-cosX)^2/(sinX)^2 Now use (sinX)^2 = 1 - (cosX)^2 to get (1-cosX)^2 / 1 - …
  2. Pre-Calc

    Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx + cosx/cosx)/(sinx/cos x - 1/cosx + cosx/cosx)= tanx + cosx (just working on the left side) ((sinx + 1 - cosx)/cosx)/((sinx …
  3. Trigonometry.

    ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated side …
  4. Trig........

    I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top and …
  5. Calculus

    I don't know if I did these problems correctly. Can you check them?
  6. Calculus

    Use the symmetry of the graphs of the sine and cosine functions as an aid in evaluating each definite integral. (a) Integral of sinx*dx from -pi/4 to pi/4 (b) Integral of cosx*dx from -pi/4 to pi/4 (c) Integral of cosx*dx from -pi/2 …
  7. Trigonometry Check

    Simplify #3: [cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] = [cosx-((1)cosx-(0)sinx)sinx]/[cosx-((-1)cosx+(0)sinx)tanx] = [cosx-cosxsinx]/[cosx+cosxtanx] = [cosx(1-sinx]/[cosx(1+tanx] …
  8. Calculus

    Evaluate the following integral integral 1 = a and b = 4 of sinx dx/(1+cos)^2 u = cosx du = -sinx dx so from here I don't know if I can do: -1 du = sinx dx or 1/sin du = x dx
  9. Calculus

    If y=3/(sinx+cosx) , find dy/dx A. 3sinx-3cosx B. 3/(sinx+cosx)^2 C. -3/(sinx+cosx)^2 D. 3(cosx-sinx)/(sinx+cosx)^2 E. 3(sinx-cosx)/(1+2sinxcosx)
  10. Calculus 2 Trigonometric Substitution

    I'm working this problem: ∫ [1-tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx) ∫cosx-sinx(sinx/cosx) ∫cosx-∫sin^2(x)/cosx sinx-∫(1-cos^2(x))/cosx sinx-∫(1/cosx)-cosx sinx-∫secx-∫cosx …

More Similar Questions