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

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Steve Share
Jan 11, 2017

Answer:

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

Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

# d/dx(uv)=u(dv)/dx+(du)/dxv #, or, # (uv)' = (du)v + u(dv) #

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

This can be extended to 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 ) :}#

Applying the product rule we get:

# 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) #

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