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

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beckden | High School Teacher | (Level 1) Educator

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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`

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