Integration Techniques, multiple pages

4.Integration with quadratics in denominator

This is the form:

$\int \frac{dx}{ax^{2}+bx +c}$

Let’s refresh ourselves with these important formulas we saw earlier:

Please note that all these are shown in videos.

$\int \frac{du}{a^{2}+u^{2}} }={\displaystyle \frac{1}{a}tan^{-1}\frac{u}{a}+C$

$\int \frac{du}{a^{2}-u^{2}} }={\displaystyle \frac{1}{2a}\ln \left | \frac{a+u}{a-u} \right |+C$

In previous problems we also discovered the following:

$\int \frac{du}{\sqrt{a^{2}-u^{2}}} }={\displaystyle \sin^{-1}\frac{u}{a}+C$

$\int \frac{du}{\sqrt{u^{2} \pm a^{2}}} }={\displaystyle \ln \left | u+\sqrt{u^{2} \pm a^{2}}\right |+C$

We will need these two formulas in this situation.

$ax^{2}+bx +c$ can be converted to one of the forms :

$u^{2} \pm a^{2}$ OR $a^{2} \pm u^{2}$ depending on the sign of $a$

That will allow us to use one of these formulas.

Problem 24
Evaluate:
$\int \frac{dx}{x^{2}+4x +9}$

We can take care of the quadratic:

$x^{2}+4x +9=(x+2)^{2}-4+9=(x+2)^{2}+5==(x+2)^{2}+(\sqrt{5})^{2}$

We apply our formula:

(1) $\begin{equation*} \begin{split} \displaystyle \int \frac{dx}{x^{2}+4x +9} }&={\displaystyle \int \frac{dx}{(x+2)^{2}+(\sqrt{5})^{2}} }\\ &={\displaystyle \frac{1}{\sqrt{5}} \tan^{-1} \frac{x+2}{ \sqrt{5}}+C} \end{split} \end{equation*}$

Finally:

$\int \frac{dx}{x^{2}+4x +9} }={\displaystyle \frac{1}{\sqrt{5}} \tan^{-1} \frac{x+2}{ \sqrt{5}}+C}={\displaystyle \frac{1}{\sqrt{5}} \arctan \frac{x+2}{ \sqrt{5}}+C$

Problem 25
Evaluate:
$\int \frac{dx}{a^{2}+x^{2}}$

We can re-write:

$\int \frac{dx}{a^{2}+x^{2}} }={\displaystyle \frac{1}{a^{2}} \int \frac{dx}{1+(\frac{x}{a})^{2}}$

Let:

$u=\frac{x}{a} \Rightarrow a\,du=dx$

$\int \frac{dx}{a^{2}+x^{2}} }={\displaystyle \frac{a}{a^{2}} \int \frac{du}{1+u^{2}} }={\displaystyle \frac{1}{a}\tan^-{1} u+C$

Back to $x$

$\int \frac{dx}{a^{2}+x^{2}} }={\displaystyle \frac{1}{a}\tan^-{1} \frac{x}{a}+C$

Finally:

$\int \frac{dx}{a^{2}+x^{2}} }={\displaystyle \frac{1}{a}\tan^-{1} \frac{x}{a}+C}={\displaystyle \frac{1}{a}\arctan \frac{x}{a}+C$

Problem 26
Evaluate:
$\int \frac{dx}{\sqrt{a^{2}-x^{2}}}$

We can re-write:

$\int \frac{dx}{\sqrt{a^{2}-x^{2}}} }={\displaystyle \frac{1}{a} \int \frac{dx}{\sqrt{1-(\frac{x}{a})^{2}}}$

Let:

$u=\frac{x}{a} \Rightarrow a\,du=dx$

$\int \frac{dx}{\sqrt{a^{2}-x^{2}}} }={\displaystyle \frac{1}{a} \int \frac{a\,du}{\sqrt{1-u^{2}}} }={\displaystyle \sin^{-1} u +C$

Back to $x$

$\int \frac{dx}{\sqrt{a^{2}-x^{2}}} }={\displaystyle \sin^{-1} \frac{x}{a} +C$

Finally:

$\int \frac{dx}{\sqrt{a^{2}-x^{2}}} }={\displaystyle \sin^{-1} \frac{x}{a} +C }={\displaystyle \arcsin \frac{x}{a} +C$

Problem 27
Evaluate: $\int \sqrt{16-x^{2}} \, dx$

This is the scenario containing $a^{2}-u^{2}$

In this istuation, $a=4$

Here $u=x$

Let $u=4 \sin \theta \Rightarrow x=4\sin \theta$

$dx=4 \cos \theta d\theta$

$\sqrt{16-x^{2}}=\sqrt{16-16\sin^{2} \theta}=\sqrt{16(1-\sin^{2} \theta)}=4\sqrt{1-\sin^{2} \theta}=4\cos \theta$

Back to the equation

(2) $\begin{equation*} \begin{split} \displaystyle \int \sqrt{16-x^{2}} \, dx }&= {\displaystyle 4\int \cos \theta \cdot 4 \cos \theta \, d\theta }\\ &= {\displaystyle 16\int \cos^{2} \theta \, d\theta }\\&={\displaystyle 16\int \frac{\cos 2\theta+1}{2} \, d\theta }\\ &={\displaystyle 8\int \cos 2\theta \, d\theta + 8\int \theta \, d\theta}\\&={\displaystyle \frac{8}{2}\sin 2\theta+8 \theta +C}\\ &= {\displaystyle 4 \cdot 2\sin \theta \cos \theta+8 \theta +C} \end{split} \end{equation*}$

But we know that:

$x=4\sin \theta \Rightarrow \sin \theta=\frac{x}{4}$

$4\cos \theta=\sqrt{16-x^{2}}\Rightarrow \cos \theta=\frac{1}{4} \sqrt{16-x^{2}}$

$x=4\sin \theta \Leftarrow \theta=\arcsin \frac{x}{4}$

Back to our original equation:

(3) $\begin{equation*} \begin{split} \displaystyle \int \sqrt{16-x^{2}} \, dx }&= {\displaystyle 4 \cdot 2\sin \theta \cos \theta+8 \theta +C}\\ &={\displaystyle 4 \cdot 2 \cdot \frac{x}{4} \cdot \frac{1}{4} \sqrt{16-x^{2}} +8 \arcsin \frac{x}{4}+C } \\ &={\displaystyle \frac{x}{2}\sqrt{16-x^{2}}+8 \arcsin \frac{x}{4}+C } \end{split} \end{equation*}$