Welcome back to Professor Baker's Math Class! In Chapter 5-5, we are tackling one of the most powerful tools in your calculus toolkit: The Substitution Rule. If you remember the Chain Rule from differential calculus, you can think of the Substitution Rule (often called u-substitution) as the Chain Rule in reverse.
1. The Core Concept: What is u-substitution?
When we face a complicated integral, we look for a composite function inside the integrand. Our goal is to simplify the problem by identifying an inner function, which we call $u$.
The formal definition states that if $u = g(x)$ is a differentiable function, then:
$$ \int f(g(x))g'(x) dx = \int f(u) du $$2. Step-by-Step: Solving Indefinite Integrals
Let's look at an example from the class notes to see this in action. Consider the integral:
$$ \int \sqrt{2x + 1} \, dx $$Step 1: Choose $u$.
We choose the inside of the square root function: $u = 2x + 1$.
Step 2: Find $du$.
Now, we differentiate $u$ with respect to $x$: $du = 2 \, dx$.
Step 3: Adjust for constants.
Looking at our original integral, we have a $dx$, but our derivative gave us $2 \, dx$. We need to isolate $dx$ to substitute it correctly. We divide by 2:
$$ \frac{1}{2} du = dx $$
Step 4: Substitute and Integrate.
Now we rewrite the integral completely in terms of $u$:
$$ \int u^{1/2} \cdot \left(\frac{1}{2}\right) du = \frac{1}{2} \int u^{1/2} du $$
Using the power rule for integration:
$$ \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right] + C = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C $$
Step 5: Back-Substitute.
Finally, replace $u$ with the original $x$-expression:
$$ \frac{1}{3}(2x+1)^{3/2} + C $$
3. Important Integrals: The Tangent Function
The notes also highlighted a classic application of substitution involving trigonometry. How do we integrate $\tan x$? We rewrite it as quotient:
$$ \int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx $$By setting $u = \cos x$ (so $du = -\sin x \, dx$), we arrive at a very useful formula:
$$ \int \tan x \, dx = \ln|\sec x| + C $$4. Definite Integrals: Changing the Limits
When dealing with definite integrals (integrals with bounds), you have two choices. However, the most efficient method is changing the limits of integration. This saves you from having to back-substitute at the end.
Example: Evaluate $\int_0^4 \sqrt{2x+1} \, dx$
- Old Limits: $x = 0$ to $x = 4$.
- Substitution: $u = 2x + 1$.
- New Lower Limit: $u(0) = 2(0) + 1 = 1$.
- New Upper Limit: $u(4) = 2(4) + 1 = 9$.
Now, substitute everything and solve using the new bounds:
$$ \frac{1}{2} \int_1^9 u^{1/2} du = \frac{1}{2} \left[ \frac{2}{3}u^{3/2} \right]_1^9 $$ $$ = \frac{1}{3} (9^{3/2} - 1^{3/2}) = \frac{1}{3}(27 - 1) = \frac{26}{3} $$Professor Baker's Tips for Success
- Pattern Recognition: Look for a function $u$ whose derivative is also present in the integral (perhaps differing only by a constant multiplier).
- Constants Matter: Don't forget to balance your constants (like the $\frac{1}{2}$ or $-\frac{1}{5}$ often seen in the notes).
- Format: For indefinite integrals, always add $+ C$. For definite integrals, changing variables is usually cleaner than back-substituting!
Keep practicing these substitutions! Once you get the hang of spotting the $u$, these problems become essentially large puzzles waiting to be solved.