What is #f(x) = int 2sin4x+tanx dx# if #f(pi/4)=1 #?

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Answer 1 · 2017-12-26T20:43:17

#f(x)=-1/2*cos(4x)-ln(cos(x))+1/2*(3-ln(2))#

#int2*sin(4x)+tan(x)dx=2intsin(4x)dx+intsin(x)/cos(x)dx#
Let #u=cos(x)# then #(du)/dx=-sin(x)#
And #s=4x# then #(ds)/dx=4#

#1/2intsin(s)ds-int1/udu#

#-1/2*cos(s)-ln(u)+C#

Substitute #u=cos(x)# and #s=4x#

#f(x)=-1/2*cos(4x)-ln(cos(x))+C#

Evaluate the constant of integration

#1=-1/2*cos(4pi/4)-ln(cos(pi/4))+C#

#C=3/2+ln(1/sqrt(2))=1/2*(3-ln(2))#

Then #f(x)=-1/2*cos(4x)-ln(cos(x))+1/2*(3-ln(2))#