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

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Answer 1 · 2018-07-18T13:24:48

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

Here ,

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

Now,

#I_1=intxsin4xdx#

#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#