How do you find #int xsin(6x) #? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer VinÃcius Ferraz Nov 21, 2015 #1 / 36 sin 6x-1/6 x cos 6x# Explanation: we want to disappear with "x" factor. #u = x, du = dx# #dv = sin 6x Rightarrow v = int sin 6x text{ d}x# #A = int u text{ d}v = xv - int v text{ d} x# #w = 6x Rightarrow frac{dw}{6} = dx# #v = int sin w (dw)/6 = - 1/6 cos w# #A = -x/6 cos 6x + int 1/6 (cos w) (dw)/6# #A = -x/6 cos 6x + 1 / 36 sin w# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1649 views around the world You can reuse this answer Creative Commons License