Section 7-3: Mastering Integration Techniques

Welcome to the recap of Professor Baker's Math Class on February 8th, 2024! Today, we dove into the world of integration, focusing on two powerful techniques: u-substitution and trigonometric substitution. Let's review the key concepts and examples we covered.

U-Substitution

U-substitution is a technique used to simplify integrals by substituting a function, $u$, for a part of the integrand. The goal is to transform the integral into a simpler form that can be easily integrated. Here's the general process:

  1. Choose a suitable $u$. Often, this is the "inner" function of a composite function or a more complicated part of the integrand.
  2. Find $du$, the derivative of $u$, with respect to $x$ (i.e., $du/dx$).
  3. Solve for $dx$ in terms of $du$ (i.e., $dx = du/(du/dx)$).
  4. Substitute $u$ and $dx$ into the original integral.
  5. Evaluate the new integral with respect to $u$.
  6. Substitute back the original expression for $u$ to get the final answer in terms of $x$.

Example: Solve $\int x \sqrt{9-x^2} dx$.

Let $u = 9 - x^2$. Then, $du = -2x dx$, so $dx = du/(-2x)$.

Substituting into the integral, we get:

$$ \int x \sqrt{u} \frac{du}{-2x} = -\frac{1}{2} \int \sqrt{u} du = -\frac{1}{2} \int u^{1/2} du $$

Now, integrate with respect to $u$:

$$ -\frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = -\frac{1}{3} u^{3/2} + C $$

Finally, substitute back $u = 9 - x^2$:

$$ -\frac{1}{3} (9-x^2)^{3/2} + C $$

Trigonometric Substitution

Trigonometric substitution is used when the integral contains expressions of the form $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$. This technique involves substituting $x$ with a trigonometric function to simplify the integral, using trigonometric identities.

Table of Trigonometric Substitutions:

  • If the integral contains $\sqrt{a^2 - x^2}$, use $x = a \sin{\theta}$.
  • If the integral contains $\sqrt{a^2 + x^2}$, use $x = a \tan{\theta}$.
  • If the integral contains $\sqrt{x^2 - a^2}$, use $x = a \sec{\theta}$.

Example: Evaluate $\int \frac{\sqrt{9-x^2}}{x^2} dx$.

Since we have $\sqrt{9 - x^2}$, we use $x = 3 \sin{\theta}$. Then, $dx = 3 \cos{\theta} d\theta$. Also, $\sqrt{9 - x^2} = \sqrt{9 - 9\sin^2{\theta}} = \sqrt{9\cos^2{\theta}} = 3\cos{\theta}$.

Substituting, we get:

$$ \int \frac{3\cos{\theta}}{9\sin^2{\theta}} 3\cos{\theta} d\theta = \int \frac{\cos^2{\theta}}{\sin^2{\theta}} d\theta = \int \cot^2{\theta} d\theta $$

Using the identity $\cot^2{\theta} = \csc^2{\theta} - 1$, we get:

$$ \int (\csc^2{\theta} - 1) d\theta = -\cot{\theta} - \theta + C $$

Now, we need to convert back to $x$. Since $x = 3\sin{\theta}$, we have $\sin{\theta} = x/3$. We can draw a right triangle where the opposite side is $x$ and the hypotenuse is $3$. The adjacent side is then $\sqrt{9 - x^2}$. Therefore, $\cot{\theta} = \frac{\sqrt{9 - x^2}}{x}$ and $\theta = \arcsin(x/3)$.

Thus, the final answer is:

$$ -\frac{\sqrt{9 - x^2}}{x} - \arcsin(\frac{x}{3}) + C $$

Keep practicing these techniques, and you'll become integration masters in no time! Remember, the key is to choose the right substitution and apply the appropriate trigonometric identities. Good luck, and see you in the next class!