Science:Math Exam Resources/Courses/MATH105/April 2014/Question 01 (i)/Solution 1

First we complete the square in the expression $3 - 2 x - x^{2}$ by considering $3 - 2 x - x^{2} = 3 - (x^{2} + 2 x) = 3 - ((x + 1)^{2} - 1) = 4 - (x + 1)^{2}$ .

This changes the integral to $\int \frac{d x}{\sqrt{3 - 2 x - x^{2}}} = \int \frac{d x}{\sqrt{4 - (x + 1)^{2}}}$

We substitute $t = x + 1$ with $d x = d t$ and obtain $\int \frac{d t}{\sqrt{4 - t^{2}}}$

Since the denominator of the last integral has the form $\sqrt{a^{2} - t^{2}}$ we can exploit trigonometric identities when we use the trigonometric substitution $t = a \sin (u)$ . In our case $a = 2$ so we substitute $t = 2 \sin (u)$ . Then $d t = 2 \cos (u) d u$ and we obtain

$\begin{aligned} \int \frac{2 \cos (u) d u}{\sqrt{4 - 4 \sin (u)^{2}}} & = \int \frac{2 \cos (u) d u}{2 \sqrt{1 - \sin (u)^{2}}} \\ = \int \frac{\cos (u) d u}{\sqrt{\cos (u)^{2}}} \end{aligned}$

where we used the trigonometric identity $1 - \sin (u)^{2} = \cos (u)^{2}$ .

The remaining integral is $\int \frac{\cos (u)}{\cos (u)} d u = \int d u = u + C$ .

Now we re-substitute $u + C = \sin^{- 1} (\frac{t}{2}) + C = \sin^{- 1} (\frac{x + 1}{2}) + C$ .