Integration Strategies

Trig Sub

For $\sqrt{b^2{x}^2-a^2}$ substitute $x=\frac{a}{b}\sec \theta$

For $\sqrt{a^2-b^2{x}^2}$ substitute $x=\frac{a}{b}\sin \theta$

For $\sqrt{a^2+b^2{x}^2}$ substitute $x=\frac{a}{b}\tan \theta$

See https://tutorial.math.lamar.edu/classes/calcii/TrigSubstitutions.aspx

Integration by Reduction Formulae

$$\int {\sec }^{n}xdx=\frac{{\sec }^{n-1}x\sin x}{n-1}+\frac{n-2}{n-1}\int {\sec }^{n-2}xdx$$

See more here: https://en.wikipedia.org/wiki/Integration_by_reduction_formulae

$u$-sub

Here's an interesting integral from Tenenbaum and Pollard's ODE book that can be solved via $u$-sub:

$$\begin{array}{cc}\int \frac{dv}{g-\frac{k{v}^2}{m}}& \text{(a)}\end{array}$$

Our strategy will be to use:

$$\begin{array}{cc}\int \frac{dv}{1-x^2}={\tanh }^{-1}x+c& \text{(b)}\end{array}$$

First factor out $g$ from the denominator and then factor the constant out of the integral:

$$\begin{array}{cc}\int \frac{dv}{g-\frac{k{v}^2}{m}}=\int \frac{dv}{g(1-\frac{k{v}^2}{mg})}=\frac{1}{g}\int \frac{dv}{1-\frac{k{v}^2}{mg}}& \text{(c)}\end{array}$$

Now we can use a $u$-sub to get the form we want. Let:

$$\begin{array}{cc}u=v\sqrt{\frac{k}{mg}},du=\sqrt{\frac{k}{mg}}dv& \text{(d)}\end{array}$$

Then:

$$\begin{array}{cc}\frac{1}{g}\int \frac{dv}{1-\frac{k{v}^2}{mg}}=\sqrt{\frac{m}{kg}}\int \frac{1}{1-\frac{k{v}^2}{mg}}\ast \sqrt{\frac{k}{mg}}dv=\sqrt{\frac{m}{kg}}\int \frac{du}{1-u^2}& \text{(e)}\end{array}$$

Now we can use the integral given in (a) and substitute back for $u$:

$$\begin{array}{cc}\sqrt{\frac{m}{kg}}\int \frac{du}{1-u^2}=\sqrt{\frac{m}{kg}}({\tanh }^{-1}u+c)=\sqrt{\frac{m}{kg}}{\tanh }^{-1}(v\sqrt{\frac{k}{mg}})+c& \text{(f)}\end{array}$$

Note that instead of using $u$-sub, we could have used partial fraction decomposition by performing different factoring to get:

$$\begin{array}{cc}\frac{-m}{k}\int \frac{dv}{(v-\frac{gm}{k})(v+\frac{gm}{k})}& \text{(g)}\end{array}$$

This would lead to a solution in terms of natural logarithm instead of ${\tanh }^{-1}$.

Integration by Parts

$$\int udv=uv-\int vdu$$