What is #F(x) = int sin3x-sinxcos^2x dx# if #F(pi) = 1 #?
1 Answer
Explanation:
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
Multiply the interior of the integral by
#=1/3intsin(3x)(3)dx-intsin(x)cos^2(x)dx#
Now that we have our
#=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
If we let the
#=-1/3cos(3x)+intcos^2(x)(-sin(x))dx#
Now, substitute our known values for
#=-1/3cos(3x)+intv^2dv#
Integrate using the rule:
#=-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
#1=(cos^3(pi)-cos(3pi))/3+C#
Note that
#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#