What is F(x) = int sin3x-sinxcos^2x dxF(x)=∫sin3x−sinxcos2xdx if F(pi) = 1 F(π)=1?

1 Answer
mason m
Mar 26, 2016

F(x)=(cos^3(x)-cos(3x)+3)/3F(x)=cos3(x)−cos(3x)+33

Explanation:

We can split this into two integrals:

F(x)=intsin(3x)dx-intsin(x)cos^2(x)dxF(x)=∫sin(3x)dx−∫sin(x)cos2(x)dx

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

Multiply the interior of the integral by 33 and the exterior by 1/313.

=1/3intsin(3x)(3)dx-intsin(x)cos^2(x)dx=13∫sin(3x)(3)dx−∫sin(x)cos2(x)dx

Now that we have our uu and dudu values present, substitute.

=1/3intsin(u)du-intsin(x)cos^2(x)dx=13∫sin(u)du−∫sin(x)cos2(x)dx

This is a common integral:

=-1/3cos(u)-intsin(x)cos^2(x)dx=−13cos(u)−∫sin(x)cos2(x)dx

=-1/3cos(3x)-intsin(x)cos^2(x)dx=−13cos(3x)−∫sin(x)cos2(x)dx

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

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

=-1/3cos(3x)+intcos^2(x)(-sin(x))dx=−13cos(3x)+∫cos2(x)(−sin(x))dx

Now, substitute our known values for vv and dvdv:

=-1/3cos(3x)+intv^2dv=−13cos(3x)+∫v2dv

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

=-1/3cos(3x)+v^3/3+C=−13cos(3x)+v33+C

=-1/3cos(3x)+cos^3(x)/3+C=−13cos(3x)+cos3(x)3+C

Combining the fractions, we see that

F(x)=(cos^3(x)-cos(3x))/3+CF(x)=cos3(x)−cos(3x)3+C

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

1=(cos^3(pi)-cos(3pi))/3+C1=cos3(π)−cos(3π)3+C

Note that cos(pi)=cos(3pi)=-1cos(π)=cos(3π)=−1.

1=((-1)^3-(-1))/3+C1=(−1)3−(−1)3+C

1=(-1+1)/3+C1=−1+13+C

1=0+C1=0+C

C=1C=1

Thus,

F(x)=(cos^3(x)-cos(3x))/3+1F(x)=cos3(x)−cos(3x)3+1

F(x)=(cos^3(x)-cos(3x)+3)/3F(x)=cos3(x)−cos(3x)+33

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