Trigonometric integration question

[turutk]

Homework Statement

\int sin^2x cos^4x dx

Homework Equations

The Attempt at a Solution

tried:
sin2x formula
writing 1-cos^2 instead of sin^2x
letting u=sinx
letting u=cosx

no luck yet

 

[lurflurf]

recall that
[sin(x)]^2[cos(x)]^4=(2+cos(2x)-2cos(4x)-cos(6x))/32
or recall that
[sin(x)]^2[cos(x)]^4=(A+B cos(2x)+C cos(4x)+D cos(6x))/32
for some numbers A,B,C,D and deduce such numbers
or integrate by parts
or make us of trigonometric reduction formula

 

[turutk]

recall that
[sin(x)]^2[cos(x)]^4=(2+cos(2x)-2cos(4x)-cos(6x))/32
or recall that
[sin(x)]^2[cos(x)]^4=(A+B cos(2x)+C cos(4x)+D cos(6x))/32
for some numbers A,B,C,D and deduce such numbers
or integrate by parts
or make us of trigonometric reduction formula

thank you for your answer. this question seems to be too long but i managed to solve it.