What is #f(x) = int x-xsecx dx# if #f((7pi)/8) = 0 #?

1 Answer
Anjali G
Mar 9, 2017

#f(x)=1/2x^2-xln|secx+tanx|-secx-3.7522#

Explanation:

#f(x)=int(x-xsecx)dx#

#f(x)=int(x)dx-color(blue)(int(xsecx)dx)#

Use integration by parts to simplify: #color(blue)(int(xsecx)dx)#
#color(blue)(u=x " " v = ln|secx+tanx|)#
#color(blue)(du=dx" "dv=secx)#
#color(blue)(int(xsecx)dx=xln|secx+tanx|-int(ln|secx+tanx|)dx)#
#color(blue)(=xln|secx+tanx|-secx+C_0)#

#f(x)=1/2x^2-xln|secx+tanx|-secx+C#

Given: #f((7pi)/8)=0#
#0=1/2((7pi)/8)^2-((7pi)/8)ln|sec((7pi)/8)+tan((7pi)/8)|-sec((7pi)/8)+C#

Since this is difficult to simplify, the rest of this response consists of approximations of #C# (using calculator).
#Capprox-3.752247019#

#f(x)=1/2x^2-xln|secx+tanx|-secx-3.7522#

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