# 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{aligned}\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{aligned}} Now we substitute:

$y={\frac {x+2}{3}}$ $dy={\frac {dx}{3}}$ $\displaystyle 3dy=dx,$ 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):

$\displaystyle y=\sin(t)$ $\displaystyle dy=\cos(t)dt$ So the above integral becomes:

{\begin{aligned}{\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{aligned}} 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{aligned}&={\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{aligned}} 