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# 𝘶-substitution

𝘶-Substitution essentially reverses the chain rule for derivatives. In other words, it helps us integrate composite functions.
When finding antiderivatives, we are basically performing "reverse differentiation." Some cases are pretty straightforward. For example, we know the derivative of start color #1fab54, x, squared, end color #1fab54 is start color #7854ab, 2, x, end color #7854ab, so integral, start color #7854ab, 2, x, end color #7854ab, d, x, equals, start color #1fab54, x, squared, end color #1fab54, plus, C. We can use this straightforward reasoning with other basic functions, like sine, left parenthesis, x, right parenthesis, e, start superscript, x, end superscript, start fraction, 1, divided by, x, end fraction, etc.
Other cases, however, are not that simple. For example, what is integral, cosine, left parenthesis, 3, x, plus, 5, right parenthesis, d, x? Hint: it's not sine, left parenthesis, 3, x, plus, 5, right parenthesis, plus, C. Try differentiating that and you will see why.
One method that can be very useful is u-substitution, which basically reverses the chain rule.

## Using $u$u-substitution with indefinite integrals

Imagine we are asked to find integral, start color #7854ab, 2, x, end color #7854ab, start color #e07d10, cosine, left parenthesis, end color #e07d10, start color #1fab54, x, squared, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, start color #7854ab, d, x, end color #7854ab. Notice that start color #7854ab, 2, x, end color #7854ab is the derivative of start color #1fab54, x, squared, end color #1fab54, which is the "inner" function in the composite function start color #e07d10, cosine, left parenthesis, end color #e07d10, start color #1fab54, x, squared, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10. In other words, letting start color #1fab54, u, left parenthesis, x, right parenthesis, equals, x, squared, end color #1fab54 and start color #e07d10, w, left parenthesis, x, right parenthesis, equals, cosine, left parenthesis, x, right parenthesis, end color #e07d10, we have:
start color #7854ab, start overbrace, 2, x, end overbrace, start superscript, u, prime, end superscript, end color #7854ab, start color #e07d10, start underbrace, cosine, left parenthesis, start color #1fab54, start overbrace, x, squared, end overbrace, start superscript, u, end superscript, end color #1fab54, right parenthesis, end underbrace, start subscript, w, end subscript, end color #e07d10, equals, start color #7854ab, u, prime, left parenthesis, x, right parenthesis, end color #7854ab, start color #e07d10, w, left parenthesis, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, right parenthesis, end color #e07d10
This suggests that u-substitution is called for. Let's see how it's done.
First, we differentiate the equation start color #1fab54, u, equals, x, squared, end color #1fab54 according to x, while treating u as an implicit function of x.
\begin{aligned} u&=x^2 \\\\ \dfrac{d}{dx}[u]&=\dfrac{d}{dx}[x^2] \\\\ \dfrac{du}{dx}&=2x \\\\ \purpleD{du}&\purpleD{=2x\,dx} \end{aligned}
In that last row we multiplied the equation by d, x so d, u is isolated. That's somewhat unorthodox, but useful for our next step. So we have start color #1fab54, u, equals, x, squared, end color #1fab54 and start color #7854ab, d, u, equals, 2, x, d, x, end color #7854ab. Now we can perform a substitution in the integral:
\begin{aligned} &\phantom{=}\displaystyle\int \purpleD{2x}\goldD{\cos(}\greenD{x^2}\goldD )\,\purpleD{dx} \\\\ &=\displaystyle\int \goldD{\cos(\greenD{\underbrace{x^2}_{u}})}\purpleD{\underbrace{2x\,dx}_{du}}&\gray{\text{Rearrange.}} \\\\ &=\displaystyle\int \goldD{\cos(}\greenD{u}\goldD )\purpleD{\,du}&\gray{\text{Substitute.}} \end{aligned}
After the substitution we are left with an expression for the antiderivative of start color #e07d10, cosine, left parenthesis, end color #e07d10, start color #1fab54, u, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10 in terms of u. How convenient! cosine, left parenthesis, u, right parenthesis is a basic function so we can find its antiderivative in a straightforward way. The only thing left to do is return the function to be in terms of x:
\begin{aligned} &\phantom{=}\displaystyle\int \goldD{\cos(}\greenD{u}\goldD )\,du \\\\ &=\sin(\greenD u)+C \\\\ &=\sin(\greenD{x^2})+C \end{aligned}
In conclusion, integral, 2, x, cosine, left parenthesis, x, squared, right parenthesis, d, x is sine, left parenthesis, x, squared, right parenthesis, plus, C. You can differentiate sine, left parenthesis, x, squared, right parenthesis, plus, C to verify that this is true.
Key takeaway #1: u-substitution is really all about reversing the chain rule:
• According to the chain rule, the derivative of start color #e07d10, w, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10 is start color #e07d10, w, prime, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, dot, start color #7854ab, u, prime, left parenthesis, x, right parenthesis, end color #7854ab.
• In u-substitution, we take an expression of the form start color #e07d10, w, prime, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, dot, start color #7854ab, u, prime, left parenthesis, x, right parenthesis, end color #7854ab and find its antiderivative start color #e07d10, w, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10.
Key takeaway #2: u-substitution helps us take a messy expression and simplify it by making the "inner" function the variable.
Problem 1.A
Problem set 1 will walk you through all the steps of finding the following integral using u-substitution.
integral, left parenthesis, 6, x, squared, right parenthesis, left parenthesis, 2, x, cubed, plus, 5, right parenthesis, start superscript, 6, end superscript, d, x, equals, question mark
How should we define u?

### Common mistake: getting incorrect expressions for $u$u or $du$d, u

Choosing the wrong expression for u will result in a wrong answer. For example, in Problem set 1, u must be defined as 2, x, cubed, plus, 5. Letting u be 6, x, squared or left parenthesis, 2, x, cubed, plus, 5, right parenthesis, start superscript, 6, end superscript will never work.
Remember: For u-substitution to apply, we must be able to write the integrand as start color #e07d10, w, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, dot, start color #7854ab, u, prime, left parenthesis, x, right parenthesis, end color #7854ab. Then, u must be defined as the inner function of the composite factor.
Another crucial step in this process is finding d, u. Make sure you are differentiating u correctly, because a wrong expression for d, u will also result in a wrong answer.
Problem 2
Tim was asked to find integral, cosine, left parenthesis, 5, x, minus, 7, right parenthesis, d, x. This is his work:
integral, cosine, left parenthesis, 5, x, minus, 7, right parenthesis, d, x, equals, sine, left parenthesis, 5, x, minus, 7, right parenthesis, plus, C
Is Tim's work correct? If not, what is his mistake?

### Common mistake: not realizing $u$u-substitution is called for

Remember: When integrating a composite function, we can't simply take the antiderivative of the outer function. We need to use u-substitution.
Letting W be an antiderivative of w, this point can be expressed mathematically as follows:
integral, w, left parenthesis, u, left parenthesis, x, right parenthesis, right parenthesis, d, x, does not equal, W, left parenthesis, u, left parenthesis, x, right parenthesis, right parenthesis, plus, C

### Another common mistake: confusing the inner function and its derivative

Imagine you're trying to find integral, x, squared, cosine, left parenthesis, 2, x, right parenthesis, d, x. You might say "since 2, x is the derivative of x, squared, we can use u-substitution." Actually, since u-substitution requires taking the derivative of the inner function, x, squared must be the derivative of 2, x for u-substitution to work. Since that's not the case, u-substitution doesn't apply here.

## Sometimes we need to multiply/divide the integral by a constant.

Imagine we are asked to find integral, start color #e07d10, sine, left parenthesis, end color #e07d10, start color #1fab54, 3, x, plus, 5, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, start color #7854ab, d, x, end color #7854ab. Notice that while we have a composite function start color #e07d10, sine, left parenthesis, end color #e07d10, start color #1fab54, 3, x, plus, 5, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, it is not multiplied by anything. That might seem weird at first, but let's proceed and see what happens.
We let start color #1fab54, u, equals, 3, x, plus, 5, end color #1fab54, then start color #7854ab, d, u, equals, 3, d, x, end color #7854ab. Now we substitute u into the integral, not before we perform this clever manipulation:
integral, start color #e07d10, sine, left parenthesis, end color #e07d10, start color #1fab54, 3, x, plus, 5, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, start color #7854ab, d, x, end color #7854ab, equals, start fraction, 1, divided by, 3, end fraction, integral, start color #e07d10, sine, left parenthesis, end color #e07d10, start color #1fab54, 3, x, plus, 5, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, start color #7854ab, 3, d, x, end color #7854ab
See what we did there? In order to have start color #7854ab, 3, d, x, end color #7854ab in the integrand, we multiplied the entire integral by start fraction, 1, divided by, 3, end fraction. That way we allowed for u-substitution while keeping the value of the integral the same.
Let's continue with the substitution:
\begin{aligned} &\phantom{=}\dfrac13\displaystyle\int\goldD{\sin(\greenD{\underbrace{3x+5}_u})}\purpleD{\underbrace{3\,dx}_{du}} \\\\ &=\dfrac13\displaystyle\int\goldD{\sin(}\greenD{u}\goldD )\purpleD{\,du} \\\\ &=-\dfrac13\cos(\greenD{u})+C \\\\ &=-\dfrac13\cos(\greenD{3x+5})+C \end{aligned}
Key takeaway: Sometimes we need to multiply or divide the entire integral by a constant, so we can achieve the appropriate form for u-substitution without changing the value of the integral.
Problem 3
integral, left parenthesis, 2, x, plus, 7, right parenthesis, cubed, d, x, equals, question mark