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

1 Answer
Dec 27, 2015

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

Explanation:

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.