What is #F(x) = int sin3x-sinxcos^2x dx# if #F(pi) = 1 #?

We can split this into two integrals:

#F(x)=intsin(3x)dx-intsin(x)cos^2(x)dx#

We will use substitution for each integral. Examining just the first, let #u=3x#, so #du=3dx#.

Multiply the interior of the integral by #3# and the exterior by #1/3#.

#=1/3intsin(3x)(3)dx-intsin(x)cos^2(x)dx#

Now that we have our #u# and #du# values present, substitute.

#=1/3intsin(u)du-intsin(x)cos^2(x)dx#

This is a common integral:

#=-1/3cos(u)-intsin(x)cos^2(x)dx#

#=-1/3cos(3x)-intsin(x)cos^2(x)dx#

For the second integral, let #v=cos(x)# and #dv=-sin(x)dx#.

If we let the #-1# on the outside of the integral in, changing the sign of the entire integral from #-# to #+#, then we will have our #dv# value:

#=-1/3cos(3x)+intcos^2(x)(-sin(x))dx#

Now, substitute our known values for #v# and #dv#:

#=-1/3cos(3x)+intv^2dv#

Integrate using the rule:#" "intv^n=v^(n+1)/(n+1)+C#

#=-1/3cos(3x)+v^3/3+C#

#=-1/3cos(3x)+cos^3(x)/3+C#

Combining the fractions, we see that

#F(x)=(cos^3(x)-cos(3x))/3+C#

We now can determine the value of the constant of integration #C# by using the original condition #F(pi)=1#.

#1=(cos^3(pi)-cos(3pi))/3+C#

Note that #cos(pi)=cos(3pi)=-1#.

#1=((-1)^3-(-1))/3+C#

#1=(-1+1)/3+C#

#1=0+C#

#C=1#

Thus,

#F(x)=(cos^3(x)-cos(3x))/3+1#

#F(x)=(cos^3(x)-cos(3x)+3)/3#