What is f(x) = int xsin4x+2tanx dx if f(pi/4)=1 ?

1 Answer
Jul 18, 2018

F(x)=1/16[sin4x-4xcos4x]+ln|sec^2x|+1-pi/16-ln2

Explanation:

Here ,

f(x)=intxsin4xdx+2int tanxdx=I_1+2I_2

Now,

I_1=intxsin4xdx

Using Integration by parts:

I_1=x*intsin4xdx-int(1*intsin4xdx)dx

=>I_1=x*(-cos(4x)/4)-int(-cos(4x)/4)dx

=>I_1=-x/4cos4x+1/4intcos4xdx

=>I_1=-x/4cos4x+1/4sin(4x)/4+c_1

=>I_1=1/16[sin4x-4xcos4x]+c_1

Again ,I_2=int tanxdx=ln|secx|+c_2

So, f(x)=I_1+2I_2

=>F(x)=1/16[sin4x-4xcos4x]+2ln|secx|+c_1+c_2

f(x)=1/16[sin4x-4xcos4x]+ln|sec^2x|+C toC=c_1+c_2

So,

f(pi/4)=1/16[sinpi-picospi]+ln|sec^2(pi/4)|+C=1

=>1/16[0-pi(-1)]+ln|(sqrt2)^2|+C=1

=>pi/16+ln2+C=1

=>C=1-pi/16-ln2

Hence ,

F(x)=1/16[sin4x-4xcos4x]+ln|sec^2x|+1-pi/16-ln2