use integration by parts to find integral of (sinx)^6 NB: (sinx)^6=(sinx)^5sinx so let u=(sinx)^5 and dv=sinxdx

so `du = 5sin^4(x)cos(x)dx` and `v =
-cos(x)`

sou

`int sin^6(x) dx
= sin^5(x) (-cos(x)) - int (-cos(x)) 5 sin^4(x)cos(x)
dx`

`int sin^6(x) dx = sin^5(x) (-cos(x)) + 5 int cos^2(x)
sin^4(x) dx`

Now we are going to use the fact that
`cos^2(x) = 1 - sin^2(x)` to get

`int sin^6(x) dx = -
sin^5(x) cos(x) + 5 int sin^4(x) dx - 5 int sin^6(x)
dx`

Now we add `5 int sin^6(x)` dx to both sides to
get

`6 int sin^6(x) dx = -sin^5(x) cos(x) + 5 int sin^4(x)
dx`

Divide both sides by 6 to
get

`int sin^6(x) dx = (-sin^5(x) cos(x))/6 + 5/6 int
sin^4(x) dx`

We do the same thing with `int sin^4(x)`
dx

Let `u = sin^3(x)` and `dv = sin(x) dx` so

`du = 3 sin^2(x) cos(x)` and `v = -cos(x)` now we
have

`int sin^4(x) dx = sin^3(x)(-cos(x)) - int (-cos(x))(3
sin^2(x) cos(x) dx` to get

`int sin^4(x) dx = -sin^3(x)
cos(x) + 3 int sin^2(x) cos^2(x) dx`

`int sin^4(x)dx =
-sin^3(x)cos(x) + 3 int sin^2(x) (1-cos^2(x))dx` so

`int
sin^4(x)dx = -sin^3(x)cos(x) + 3 int sin^2(x) dx - 3 int sin^4(x))`
dx

Now adding `3 int sin^4(x) dx` to both
sides

`4 int sin^4(x)dx = -sin^3(x)cos(x) + 3 int sin^2(x)
dx`

Divide both sides by 4 to
get

`int sin^4(x)dx = (-sin^3(x)cos(x))/4 + 3/4 int
sin^2(x) dx`

We can do the same thing with `int sin^2(x)
dx`

to get

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

Add `int
sin^2(x) dx` and solving to get

` int sin^2(x) dx =
(-sin(x)cos(x) + x)/2 + C`

Now we substitute
back into the `int sin^4(x) dx` to get

`int sin^4(x)dx =
-(sin^3(x)cos(x))/4 + 3/4(-sin(x)cos(x) + x)/2 + C`

and
substitute back into int sin^6(x) dx and simplifying we
get

`int sin^6(x) dx = (-sin^5(x) cos(x))/6 + 5/6
(-(sin^3(x)cos(x))/4 + 3/4(-sin(x)cos(x)+x)/2) +
C'`

Simplifying we get the final
answer:

`int sin^6(x)dx = -((cos
xsin⁵x)/6)-(5/(24))cos xsin³x`

-(5/(16))cos xsin
x-(5/(16))x + C`