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