Welcome back, class! We have a few important administrative updates regarding our schedule and the upcoming test, followed by detailed review notes covering the questions you asked during last week's lesson.

Class Schedule & Test Update

To ensure everyone has a solid grasp of the material, we are making a strategic adjustment to the schedule:

  • Sections 7-7 and 7-8: Rather than relying solely on videos, we will cover these crucial topics directly in class next week. These lessons are foundational for future math courses, so I want to ensure we work through them together to guarantee a strong understanding.
  • Test Date: As a result of this shift, the test will not be held on the 16th. We will discuss the new date in class.

Review: Integration by Parts (Section 7.1)

Below are clarifications on specific homework problems using the Integration by Parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Problem #1: $\int x e^{2x} \, dx$

For this problem, we follow the LIATE rule to select our $u$. Since $x$ is Algebraic and $e^{2x}$ is Exponential, we set:

  • $u = x \rightarrow du = dx$
  • $dv = e^{2x} dx \rightarrow v = \frac{1}{2}e^{2x}$

Plugging this into our formula gives us:

$$ = x(\frac{1}{2}e^{2x}) - \int \frac{1}{2}e^{2x} \, dx $$ $$ = \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + C $$

Problem #29: $\int_0^{\pi} x \sin x \cos x \, dx$

Pro Tip: Before rushing into Integration by Parts, simplify the integrand using trigonometric identities. Recall that $\sin x \cos x = \frac{1}{2}\sin(2x)$.

This simplifies the integral to:

$$ \frac{1}{2} \int_0^{\pi} x \sin(2x) \, dx $$

Now, the integration by parts is much cleaner with $u=x$ and $dv=\sin(2x)dx$.

Review: Trigonometric Integrals (Section 7.2)

Problem #3: $\int_0^{\pi/2} \sin^7 \theta \cos^5 \theta \, d\theta$

When dealing with products of powers of sine and cosine, if the power of cosine is odd (in this case, 5), the strategy is to save one cosine factor to be part of your $du$ and express the remaining factors in terms of sine.

We rewrite $\cos^5 \theta$ as $\cos^4 \theta \cdot \cos \theta = (1-\sin^2 \theta)^2 \cos \theta$.

Using $u$-substitution:

  • $u = \sin \theta$
  • $du = \cos \theta \, d\theta$
  • Change limits: $u(0)=0$ and $u(\pi/2)=1$

The integral becomes a polynomial which is much easier to solve:

$$ \int_0^1 u^7 (1-u^2)^2 \, du = \frac{1}{120} $$

Please review the attached handwritten notes for the full step-by-step work for these problems and others (including Problem #13 and #25). Keep up the hard work!