Science:Math Exam Resources/Courses/MATH101/April 2005/Question 03 (c)/Solution 1

We want to integrate:

$\int \frac{dx}{(5-4x-x^2{)}^{3/2}}.$

The first step is to complete the square in the denominator:

$\begin{array}{rl}\int \frac{dx}{(9-4-4x-x^2{)}^{3/2}}& =\int \frac{dx}{(9-(x+2{)}^2{)}^{3/2}}\\ & =\int \frac{1}{27}\frac{dx}{(1-\frac{(x+2{)}^2}{9}{)}^{3/2}}\\ & =\frac{1}{27}\int \frac{dx}{(1-(\frac{x+2}{3}{)}^2{)}^{3/2}}\end{array}$

Now we substitute:

$y=\frac{x+2}{3}$

$dy=\frac{dx}{3}$

$3dy=dx,$

So the integral simplifies to

$\frac{3}{27}\int \frac{dy}{(1-y^2{)}^{3/2}}=\frac{1}{9}\int \frac{dy}{(1-y^2{)}^{3/2}}$

We now perform a trigonometric substitution (this step is only valid for |y| smaller than 1, which we know is true because 1-y^2 must be positive):

$y=\sin (t)$

$dy=\cos (t)dt$

So the above integral becomes:

$\begin{array}{rl}\frac{1}{9}\int \frac{\cos (t)dt}{(1-{\sin }^2(t){)}^{3/2}}& =\frac{1}{9}\int \frac{\cos (t)dt}{({\cos }^2(t){)}^{3/2})}\\ & =\frac{1}{9}\int \frac{dt}{{\cos }^2(t)}\\ & =\frac{1}{9}\int {\sec }^2(t)dt\\ & =\frac{1}{9}\tan (t)+C\\ & =\frac{1}{9}\tan (\arcsin (y))+C\\ & =\frac{1}{9}\tan (\arcsin (\frac{x+2}{3}))+C\end{array}$

But for any u, we have

$\tan (\arcsin (u))=\frac{u}{\sqrt{1-u^2}},$

So the above quantity is equal to

$=\frac{1}{9}\frac{\frac{x+2}{3}}{\sqrt{1-(\frac{x+2}{3}{)}^2}}$

$=\frac{1}{9}\frac{x+2}{\sqrt{5-4x-x^2}}.$

So we have

$\int \frac{dx}{(5-4x-x^2{)}^{3/2}}=\frac{1}{9}\frac{x+2}{\sqrt{5-4x-x^2}}.$

This is not required in a test, but we can check our answer by differentiating:

$\frac{d}{dx}\frac{1}{9}\frac{x+2}{\sqrt{5-4x-x^2}}$

$\begin{array}{rl}& =\frac{1}{9}\frac{\sqrt{5-4x-x^2}-\frac{1}{2}(x+2)(5-4x-x^2{)}^{-1/2}(-2x-4)}{(5-4x-x^2)}\\ & =\frac{1}{9}\frac{5-4x-x^2+x^2+2x+2x+4}{(5-4x-x^2{)}^{3/2}}\\ & =\frac{1}{9}\frac{9}{(5-4x-x^2{)}^{3/2}}\\ & =\frac{1}{(5-4x-x^2{)}^{3/2}}\end{array}$