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

1 Answer
Mar 26, 2016

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

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 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