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

1 Answer
Dec 26, 2017

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

Explanation:

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))