Hannah

Lesson Plan: 8th August

Lesson Plan: 1st August

Lesson Plan: 25th July

Integration by substitution

We’d cover integration by substitution first. At the A Levels, any integration that needs to be done by substitution will have the substitution given to us. We can then follow these next general steps (we will assume that $x$ is our original variable and $u$ is our new variable.

  1. Differentiate the given substitution (to change the “$\mathrm{d}x$” into “$\mathrm{d}u$”.
  2. Substitute both $x$ and “$\mathrm{d}x$” into the “$u$”-counterparts.
  3. After substitution and simplification, we should now have an integral that can be done relatively simply.
  4. Change all $u$s back to the original variables $x$’s.

Example 1

20d) Use the substitution $u=e^{2x}$ to find $\displaystyle \int \frac{e^{2x}}{\sqrt{1-e^{4x}}} , \mathrm{d}x$.

Step 1: $\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x} = 2 e^{2x}$

“$\displaystyle \mathrm{d}x = \frac{1}{2e^{2x}} , \mathrm{d}u$”

Step 2:

$\displaystyle \int \frac{e^{2x}}{\sqrt{1-e^{4x}}} , \mathrm{d}x =\int \frac{u}{\sqrt{1-u^2}} \left( \frac{1}{2u} , \mathrm{d}u\right)$

Step 3:
$\displaystyle = \frac{1}{2} \int \frac{1}{\sqrt{1-u^2}} , \mathrm{d}u = \frac{1}{2} \sin^{-1} u + c$

Step 4:
$\displaystyle = \frac{1}{2} \sin^{-1} (e^{2x}) +c$

Example 2:

For trigo substitutions there may be a few extra tricks, such as the use of our Pythagorean/square formulas, and using the right angled triangle to simplify certain steps (in step 4 in this example: I’d draw it out for you during class).

Modified 17e) Use the substitution $x = \sec \theta$ to find $\displaystyle \int \frac{1}{\sqrt{x^2-1}} , \mathrm{d}x$.

Step 1: $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}\theta} = \sec \theta \tan \theta$

“$\displaystyle \mathrm{d}x = \sec \theta \tan \theta , \mathrm{d}\theta$”

Step 2:

$\displaystyle \int \frac{1}{\sqrt{x^2-1}} , \mathrm{d}x =\int \frac{1}{\sqrt{\sec^2 \theta – 1}} \left( \sec \theta \tan \theta , \mathrm{d}\theta \right)$

Step 3:
$\displaystyle =\int \frac{1}{\sqrt{\tan^2 \theta}} \sec \theta \tan \theta , \mathrm{d}\theta = \int \sec \theta , \mathrm{d}\theta = \ln | \sec \theta + \tan \theta |+ c$

Step 4:
$\displaystyle = \ln | x + \sqrt{x^2-1} | +c$

Integration by parts

$\displaystyle \int u \frac{\mathrm{d}v}{\mathrm{d}x} , \mathrm{d}x = uv – \int \frac{\mathrm{d}u}{\mathrm{d}x}v,\mathrm{d}x$

Integration by parts is analogous to “product” rule in differentiation, except that the formula is a lot more complicated, and there is still an integration sign left after applying the formula.

Successfully applying integration by parts involve making the second term simpler such that we can now integrate it.

Example 1

14d) $\displaystyle \int x \cos x , \mathrm{d}x$

$\displaystyle \int x \cos x , \mathrm{d}x = x \sin x – \int 1 \sin x , \mathrm{d}x = x \sin x + \cos x + c$

A discussion on order

We will discuss more about order in class, and how it is important in this technique.
If no hints are given, a useful heuristic is LIATE

  • L: Logarithm: $\ln x$
  • I: Inverse Trigo: $\sin^{-1}x, \tan^{-1}x$
  • A: Algebra: $1, x, x^2$
  • T: Trigo: $\sin x, \cos x$
  • E: Exponential

Roughly we group them into 3:

  • LI: Almost always ‘differentiated’ (i.e. the ‘$u$’)
  • TE: Almost always ‘integrated’ (i.e. the ‘$\frac{\mathrm{d}v}{\mathrm{d}x}$’)
  • A: KIV: depending on who they’re paird with