Integral of xsin^2(6x)dx ?

Question

I have seen integration by parts of xsin^2x but would like to see if using trigonometric identities would streamline this problem?

Answer 1

no, you can't use trig sub unless you're dealing with an integrant that has the form √(x²-a²), √(a²-x²) or √(x²+a²)

sin²(6x) = 1/2 - (1/2)cos(12x)

distribute and the integral becomes:

∫(1/2)x - (1/2)x cos(12x)

the first term is straight forward. (1/4)x²

To integrate the second term, use parts:

∫u dv = uv - ∫v du

let u = (1/2)x ; dv = cos(12x)

du = (1/2)dx ; v = (1/12)sin(12x)

substitute:

(1/24)x sin(12x) - ∫ (1/24)sin(12x) dx

(1/124)x sin(12x) + (1/288)cos(12x)

so you have:

(1/4)x² - [(1/124)x sin(12x) + (1/288)cos(12x)]

distribute:

(1/4)x² - (1/124)x sin(12x) - (1/288)cos(12x) + C

hope it helps!