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

Question

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

William Abdalla · Accepted Answer

#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))#

William Abdalla · Answer

undefined To find $ f(x) = \int (2\sin(4x) + \tan(x)) \, dx $ when $ f\left(\frac{\pi}{4}\right) = 1 $, first integrate the function with respect to $ x $:

$$ f(x) = \int (2\sin(4x) + \tan(x)) \, dx $$

$$ = -\frac{1}{2}\cos(4x) + \ln|\sec(x)| + C $$

Then, use the given condition $ f\left(\frac{\pi}{4}\right) = 1 $ to find the constant $ C $:

$$ 1 = -\frac{1}{2}\cos\left(\frac{\pi}{4}\right) + \ln|\sec\left(\frac{\pi}{4}\right)| + C $$

$$ 1 = -\frac{1}{2}\left(\frac{\sqrt{2}}{2}\right) + \ln\left|\sec\left(\frac{\pi}{4}\right)\right| + C $$

$$ 1 = -\frac{\sqrt{2}}{4} + \ln\left|\sqrt{2}\right| + C $$

$$ 1 = -\frac{\sqrt{2}}{4} + \frac{1}{2}\ln(2) + C $$

$$ C = 1 + \frac{\sqrt{2}}{4} - \frac{1}{2}\ln(2) $$

Now, substitute $ C $ back into the expression for $ f(x) $:

$$ f(x) = -\frac{1}{2}\cos(4x) + \ln|\sec(x)| + 1 + \frac{\sqrt{2}}{4} - \frac{1}{2}\ln(2) $$

This is the expression for $ f(x) $.