How do you integrate #f(x)=e^xsinxcosx# using the product rule?

Question

Christopher Allers · Accepted Answer

# f'(x) = e^x(sinxcosx + cos^2x - sin^2x) #

The Product Rule for Differentiation is something you should learn and practice using if you are studying math: # d/dx(uv)=u(dv)/dx+(du)/dxv #, or, # (uv)' = (du)v + u(dv) # "The first times the derivative of the second plus the derivative of the first times the second" is the rule that I was taught to memorize. This applies to the following three products: # d/dx(uvw)=uv(dw)/dx+u(dv)/dxw + (du)/dxvw# So with # f(x) = e^xsinxcosx # we have; # { ("Let "u = e^x, => , (du)/dx = e^x), ("And "v=sinx, =>, (dv)/dx = cosx ), ("And "w = cosx, =>, (dw)/dx = -sinx ) :}# When we use the product rule, we obtain: # d/dx(uvw) = (du)/dxvw + u(dv)/dxw+ + uv(dw)/dx# # :. f'(x) = (e^x)(sinx)(cosx) + (e^x)(cosx)(cosx) + (e^x)(sinx)(-sinx) # # " " = e^x(sinxcosx + cos^2x - sin^2x) #

Christopher Allers · Answer

undefined To integrate $ f(x) = e^x \sin(x) \cos(x) $ using the product rule, you first express it as a product of two functions: $ u = e^x $ and $ v = \sin(x)\cos(x) $. Then, you apply the product rule for differentiation, which states that the derivative of the product of two functions $ u $ and $ v $ is given by $ u'v + uv' $.

Differentiating $ u = e^x $ with respect to $ x $ gives $ u' = e^x $, and differentiating $ v = \sin(x)\cos(x) $ gives $ v' = \cos^2(x) - \sin^2(x) $.

Now, you integrate $ u'v + uv' $ with respect to $ x $, which yields the integral of $ e^x (\cos^2(x) - \sin^2(x)) + e^x \sin(x)\cos(x) $.

Finally, you integrate each term separately using the appropriate techniques for integration.