Question

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

Answer 1

`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.