Question

How do you find the integral of `sin^2(x)cos^4(x)`?

Answer 1

`int sin^2(x)cos^4(x) dx`

`= -1/6 sin^5(x) cos(x)+7/24 sin^3(x) cos(x)-1/16 sin(x) cos(x)+1/16 x + C`

Explanation:

First note that:

`sin^2(x)cos^4(x) = (1 - cos^2(x)) cos^4(x) = -cos^(6)x + cos^4(x)`

`d/(dx) sin^5(x) cos(x) = 5 sin^4(x) cos^2(x) - sin^6(x)`

`color(white)(d/(dx) sin^5(x) cos(x)) = sin^4(x)(5 cos^2(x) - sin^2(x))`

`color(white)(d/(dx) sin^5(x) cos(x)) = (1-cos^2(x))(1-cos^2(x))(5 cos^2(x) - (1-cos^2(x)))`

`color(white)(d/(dx) sin^5(x) cos(x)) = (cos^4(x)-2cos^2(x)+1)(6 cos^2(x) - 1)`

`color(white)(d/(dx) sin^5(x) cos(x)) = 6cos^6(x)-13cos^4(x)+8cos^2(x)-1`

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

`color(white)(d/(dx) sin^3(x) cos(x)) = sin^2(x)(3 cos^2(x) - sin^2(x))`

`color(white)(d/(dx) sin^3(x) cos(x)) = (1-cos^2(x))(3 cos^2(x) - (1-cos^2(x)))`

`color(white)(d/(dx) sin^3(x) cos(x)) = (-cos^2(x)+1)(4 cos^2(x) - 1)`

`color(white)(d/(dx) sin^3(x) cos(x)) = -4cos^4(x)+5cos^2(x)-1`

`d/(dx) sin(x) cos(x) = cos^2(x)-sin^2(x) = 2cos^2(x) - 1`

Now we can choose multipliers to make the running sum match each coefficient of `cos^(2k)(x)` in descending order:

`d/(dx) (-1/6 sin^5(x) cos(x)) = -cos^6(x)+13/6 cos^4(x)-4/3 cos^2(x)+1/6`

`d/(dx) (7/24 sin^3(x) cos(x)) = -7/6 cos^4(x)+35/24 cos^2(x) - 7/24`

`d/(dx) (-1/16 sin(x) cos(x)) = -1/8 cos^2(x) + 1/16`

`d/(dx) (1/16 x) = 1/16`

summing to: `-cos^6(x)+cos^4(x) = sin^2(x)cos^4(x)`

So:

`int sin^2(x)cos^4(x) dx`

`= -1/6 sin^5(x) cos(x)+7/24 sin^3(x) cos(x)-1/16 sin(x) cos(x)+1/16 x + C`