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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

$$\sqrt{4-x^2} - \sqrt{5} \tan ^{-1} \left ( \frac{3x+4}{\sqrt{5}\sqrt{4-x^2}}\right ) + 3 \sin ^{-1} \left ( \frac{x}{2} \right )+C $$

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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 $\cot y=u$

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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.

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    $\begingroup$ OP already used this substitution... $\endgroup$ – Andres Mejia Apr 13 '16 at 13:05
  • $\begingroup$ Correct. However I suspect OP may have not kept on substituting. The point my answer makes is that the integral can be calculated by repeated substitution. The OP also mentioned that the context of the question is that only elementary methods should be used. -EDIT: I reread the question. You are correct. My mistake. $\endgroup$ – Tomi Apr 13 '16 at 13:08

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