What is #f(x) = int xtanx dx# if #f(pi/4)=-2 #?

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Answer 1 · 2015-12-27T07:11:52

#f(x) = int_{pi/4}^x ln(cosu) du - xln(cosx) - frac{piln2}{8} - 2#

#f'(x) = x*tan(x)#

#f(x) - f(pi/4) = int_{pi/4}^x u*tan(u) du#

#f(x) - (-2) = int_{pi/4}^x u*tan(u) du#

#f(x) = int_{pi/4}^x u*tan(u) du - 2#

Try to evaluate the integral using integration by parts.

#int_{pi/4}^x u*tan(u) du = -int_{pi/4}^x u*frac{d}{du}(ln(cosu)) du#

#= -( [u*ln(cosu)]\_{pi/4}^x - int_{pi/4}^x ln(cosu)*frac{d}{du}(u) du )#

#= - ( xln(cosx) - pi/4ln(1/sqrt{2}) ) + int_{pi/4}^x ln(cosu) du #

#= - xln(cosx) - frac{piln2}{8} + int_{pi/4}^x ln(cosu) du#

Evaluating the integral fully requires non-elementary functions.