Question

What is the integral of `sin^3x`?

Answer 1

`1/12(cos3x-9cosx)+C`.

Explanation:

We know that, `sin3x=3sinx-4sin^3x rArr sin^3x=1/4(3sinx-sin3x)`.

`:. intsin^3xdx`,

`=1/4int(3sinx-sin3x)dx`,

`=1/4{3(-cosx)-(-cos3x)1/3}`.

`rArr intsin^3xdx=1/12(cos3x-9cosx)+C`.

Answer 2

`int sin^3 xcolor(white)(.)dx = -cos x + 1/3 cos^3 x + C`

Explanation:

As is often the case with trigonometric integrals, there is more than one way to tackle this...

Note that:

`sin^2 x = 1-cos^2 x`

So:

`int sin^3 x dx = int sin x(1 - cos^2 x)dx`

`color(white)(int sin^3 x dx) = int sin x - sin x cos^2 x color(white)(.)dx`

`color(white)(int sin^3 x dx) = -cos x + 1/3 cos^3 x + C`

Note that this is equivalent to the other answer, as you may care to verify yourself. [Hint: Use `cos 3 x = 4 cos^3x - 3 cos x`]