Question

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

Answer 1

`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))`