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

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

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

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

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 $dt=2\cos(u)du$ and we obtain

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

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

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

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