Find $\int_{-1}^{1} \frac{\sqrt{4-x^2}}{3+x}dx$ I came across the integral $$\int_{-1}^{1} \frac{\sqrt{4-x^2}}{3+x}dx$$ in a calculus textbook. At this point in the book, only u-substitutions were covered, which brings me to think that there is a clever substitution that one can use to knock off this integral.
I was able to find the answer using $x= 2 \sin \theta$, doing a bit of polynomial long division and then Weiestrass substitution. However, this solution was rather ugly and I don't think this was what the author had in mind.
What else could I try here? Wolfram gives a closed form $$\pi + \sqrt{5} \left [ \tan ^{-1} \left ( \frac{1}{\sqrt{15}} \right ) - \tan ^{-1} \left ( \frac{7}{\sqrt{15}} \right ) \right ]$$
and the indefinite integral was 
$$\sqrt{4-x^2} - \sqrt{5} \tan ^{-1} \left ( \frac{3x+4}{\sqrt{5}\sqrt{4-x^2}}\right ) + 3 \sin ^{-1} \left ( \frac{x}{2} \right )+C $$
 A: HINT:
Using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$
$$I+I=6\int_{-1}^1\dfrac{\sqrt{4-x^2}}{9-x^2}dx$$
Putting $x=2\sin y\implies$
$$2I=6\int_{-\pi/6}^{\pi/6}\dfrac{4\cos^2y}{9-4\sin^2y}dy$$
$$=6\int_{-\pi/6}^{\pi/6}\dfrac{5+4(1-\sin^2y)-5}{9-4\sin^2y}dy$$
$$=6\int_{-\pi/6}^{\pi/6}\left(1-\dfrac5{9-\sin^2y}\right)dy$$
Now, $$\int\dfrac{dy}{9-\sin^2y}=\int\dfrac{\csc^2y}{9(1+\cot^2y)-1}dy$$
Set $\cot y=u$
A: One option is to substitute 
$$x=2\sin(u)$$
$$dx=2\cos(u)du $$
This means our integral is now 
$$=-2 \int _{-\pi/6}^{\pi/6} \frac{2 \cos(u) (\cos^2(u)^{1/2})}{2\sin(u)+3} $$
If we simplify 
$$4 \int _{-\pi/6}^{\pi/6} \frac{(\cos^2(u))}{2\sin(u)+3} $$
Now we carry out another substitution of 
$$s=\tan(u/2)$$
$$ds=1/2 \sec^2(u/2) du$$
This will give the "more" manageable integral of  
$$4 \int _{-3^{(1/2)}-2}^{2-(3)^{(1/2)}} \frac{(s^2-1)^2)}{(s^2+1)^2(3s^2+4s+3)}$$
We can now use partial fractions which gives 3 functions (one which requires another substitution) two of which are the correct integral for $\tan^{-1}$ and one which is odd. Which should now give 
$$=\pi + (5)^{(1/2)}(\tan^{-1}(1/(15^{(1/2)})-\tan^{-1}(7/(15^{(1/2)}))$$
That you said wolfram agreed with. 
