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##

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##