Question

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

Answer 1

`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`