sin² x cos⁴ x should be changed first into something with no powers, if possible.


∫ sin² x cos⁴ x dx

= ∫ sin² x cos² x cos² x

= ∫ (sin x cos x)² (½(cos 2x + 1)) dx

[Recall double angle formula:
cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A
So, 2cos² A = cos 2A + 1
cos² A = ½(cos 2A + 1)
Simply replace A with x]

= ∫ (½ sin 2x)² (½(cos 2x + 1))

[Recall double angle formula:
sin 2A = 2 sin A cos A
So , sin A cos A = ½ sin 2A
Likewise, replace A with x ]


= ∫ ¼ sin² 2x (½(cos 2x + 1)) dx

= ∫ [⅛ sin² 2x cos 2x + ⅛ sin² 2x] dx

The expression on the left can be rewritten such that it is directly integrable. The expression on the right can be changed using double angle formula.

= ∫ [ 1/48 (3(sin² 2x)(2cos 2x)) + ⅛ (½(1 - cos 4x)) ] dx

= 1/48 sin³ 2x + 1/16 (x - ¼ sin 4x)

= 1/48 sin³ 2x + 1/16 x - 1/64 sin 4x + c

Alternatively,


∫ sin² x cos⁴ x dx

= ∫ sin² x (cos² x)² dx

= ∫ ½(1 - cos 2x) (½(1 + cos 2x))² dx

= ∫ ⅛ (1 - cos 2x)(1 + cos 2x)(1 + cos 2x) dx

= ∫ ⅛ (1 - cos² 2x)(1 + cos 2x) dx

[Recall (a + b)(a - b) = a² - b²]

= ∫ ⅛ (sin² 2x)(1 + cos 2x) dx

= ∫ ⅛ (sin² 2x + sin² 2x cos 2x) dx

= ∫ ⅛ (½(1 - cos 4x) + 1/6 (3(sin 2x)² (2 cos 2x) ) ) dx

= ⅛ ( ½(x - ¼ sin 4x) + 1/6 sin³ 2x)

= 1/16 x - 1/64 sin 4x + 1/48 sin³ 2x + c