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December 23, 2012
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# Integrate xcotx having limit 0 to pi/2?

by gôhpihán
Member since:
March 29, 2009
Total points:
41,534 (Level 7)

## Best Answer - Chosen by Voters

Integrate by Parts

∫(0 to π/2) x cotx dx, let u = x, dv = cotx dx ==> du = dx, v = ln|sinx| dx

= ∫(0 to π/2) u dv

= uv - ∫(0 to π/2) v du

= x ln|sinx| - ∫(0 to π/2) ln|sinx| dx

= 0 - ∫(0 to π/2) ln|sinx| dx, because evaluating at x approaches 0 and pi/2 of x ln|sinx| yields 0

= - ∫(0 to π/2) ln|sinx| dx

Let I = -∫(0 to π/2) ln|sinx| dx, let u = π/2 - x

I = -∫(0 to π/2) ln|cosu| du

I = -∫(0 to π/2) ln|cosx| dx, dummy variable

2I = -∫(0 to π/2) ln|sinx cosx| dx

2I = -∫(0 to π/2) ln|sin2x| dx + (π/2) ln2

2I = -(1/2) ∫(0 to π) ln|sinu| du + (π/2) ln2

-4I = ∫(0 to π) ln|sinu| du - πln2

-4I = ∫(0 to π/2) ln|sinu| du + ∫(π/2 to π) ln|sinu| du - πln2

-4I = -I + ∫(π/2 to π) ln|sinu| du - πln2

-3I = ∫(π/2 to π) ln|sinu| du - πln2, v = u - π/2

-3I = ∫(0 to π/2) ln|cosv| du - πln2

-3I = -I - πln2

2I = π ln2

I = (π/2) ln2

Confirmation:
http://www.wolframalpha.com/input/?i=int…
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• by shahraiz
Member since:
September 07, 2012
Total points:
288 (Level 2)
use integration by parts x as first and cotx as 2nd:)