When we are reversing a differentiation that had the composition of three functions. Here is one example.
##int sin^4(7x)cos(7x)dx##
Let ##u=7x##. This makes ##du = 7dx## and our integral can be rewritten:
##1/7 int sin^4ucosudu = 1/7int(sinu)^4cosudu##
To avoid using ##u## to mean two different things in one discussion, we’ll use another variable (##t, v, w## are all popular choices)
Let ##w=sinu##, so we have ##dw = cosudu## and our integral becomes:
##1/7intw^4dw##
We the integrate and back-substitute:
##1/7intw^4dw = 1/35 w^5 +C##
## = 1/35 sin^5u +C##
## = 1/35 sin^5 7x +C##
If we check the answer by differentiating, we’ll use the twice.
##d/dx((sin(7x))^5) = 5(sin(7x))^4*d/dx(sin(7x))##
## = 5(sin(7x))^4*cos(7x)d/dx(7x)##
## = 5(sin(7x))^4*cos(7x)*7##
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