Lecture Notes on Integration Techniques

Integral of Polynomial Functions

• We are trying to find the integral of $x \cdot \cos(x) dx$.
• Recognize that there is a product of two functions, $x$ and $\cos(x)$.
• The problem suggests using u-substitution as the primary method for solving integrals.

U-Substitution

• The method applies to integrals such as $\int x \cos(x) dx$ or $\int \sin(t^2) \cos(t^2) t dt$.

Example 1: Integral of $\sin(t) \cos(t)$

• Consider the integral $\int \sin(t) \cos(t) dt$.
• Here you have a function and its derivative.
• Let $u = \sin(t)$. Then $\frac{du}{dt} = \cos(t)$, so $du = \cos(t) dt$.
• The integral becomes $\int u du = \frac{1}{2}u^2 + c = \frac{1}{2} \sin^2(t) + c$.
• The integral of $\sin(t)$ is $-\cos(t) + c$.

Example 2: Integral of $\sin(t^2) \cos(t^2) t dt$

• Consider the integral $\int \sin(t^2)\cos(t^2)t dt$.
• Let $u = \sin(t^2)$.
• Then $\frac{du}{dt} = \cos(t^2) \cdot 2t$, so $\frac{1}{2} du = t \cos(t^2) dt$.
• The integral becomes $\int \frac{1}{2} \sin(t^2) du = \frac{1}{2} \int u du = \frac{1}{2} \cdot \frac{u^2}{2} + c = \frac{1}{4}u^2 + c$.
• Substituting back, we get $\frac{1}{4}\sin^2(t^2) + c$.
• In this example, we can use u-substitution because the derivative of $\sin(t^2)$ is present in the integral.

Derivatives of Composite Functions

• The derivative of $\sin(t^2)$ is $\cos(t^2) \cdot 2t$.
• Using the power rule and chain rule, the derivative of $\sin^2(t^2)$ is $2 \sin(t^2) \cdot \cos(t^2) \cdot 2t = 4t \sin(t^2) \cos(t^2)$.

Example 3: Integral of $t \sin(t^2) \cos(t^2) dt$

• We need to solve $\int t \sin(t^2) \cos(t^2) dt$.
• Let $u = \sin(t^2)$. Then $\frac{du}{dt} = \cos(t^2) \cdot 2t$, so $du = 2t \cos(t^2) dt$.
• Solve for $dt$ to get $\frac{du}{2t \cos(t^2)} = dt$.
• Substituting, we have $\int t u \cos(t^2) \frac{du}{2t \cos(t^2)}$.
• Which simplifies to $\int \frac{1}{2} u du = \frac{1}{2} \frac{u^2}{2} + c = \frac{1}{4} u^2 + c = \frac{1}{4} \sin^2(t^2) + c$.

Practice Problems

• The lecture mentions providing 10 practice problems.
• Example problem: $\int e^x \sqrt{1 + e^x} dx$
• Let $u = 1 + e^x$. Then $du = e^x dx$.
• So the integral becomes $\int \sqrt{u} du$.
• Another example: $\int e^{x^2} x dx$
• Let $u = x^2$. So $du = 2x dx$.
• This implies $\frac{1}{2} du = x dx$.
• Replace $x^2$ with $u$ in the exponential, then replace $x dx$ with $\frac{1}{2} du$.
• New integral: $\int e^u \frac{1}{2} du$.

Importance of U-Substitution from Previous Classes

• U-substitution is a fundamental concept.
• Advanced integration techniques may require multiple pages to solve, while u-substitution provides a more straightforward approach.
• Understanding derivative rules is essential for mastering anti-derivatives.

Dealing with Derivatives and Anti-derivatives

• Review derivative rules to better understand anti-derivatives.
• The lecture states that you need to understand the derivative rules in order to apply anti derivative rules.

Formula Sheet

• A formula sheet may not be provided for the exam.

Example 4: Integral of $e^{\frac{1}{x}} \frac{1}{x^2} dx$

• Let $u = \frac{1}{x}$, so $u = x^{-1}$.
• $\frac{du}{dx} = -1 \cdot x^{-2} = -\frac{1}{x^2}$.
• $du = -\frac{1}{x^2} dx\rightarrow -du = \frac{1}{x^2} dx$.
• Substitute back into the integral: $\int e^u (-1) du = - \int e^u du = -e^u + c$.
• Substituting back for u: $-e^{\frac{1}{x}} + c$.
• To verify, find the derivative of $f(x) = -e^{\frac{1}{x}} + c$.
• $f'(x) = -e^{\frac{1}{x}} \cdot (-\frac{1}{x^2}) + 0 = \frac{e^{\frac{1}{x}}}{x^2}$, which confirms the anti-derivative.