# How to compute $\int\frac{x^2}{x^2+4}\,dx$ and $\int\frac{1}{\sqrt{4-x^2}}\,dx$?

My calculus teacher gave us a list of integrals to solve through the substitution method. I've been trying everything for hours, and I just can't find out how to compute these: $$∫\frac{x^2}{x^2+4}\,dx\qquad\text{and}\qquad \int\frac{1}{\sqrt{4-x^2}}\,dx.$$

Is it even possible to solve them through substitution?

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The first one can be solved by rewriting it first $$\int\frac{x^2}{x^2+4}\,dx = \int\frac{x^2+4-4}{x^2+4}\,dx = \int\,dx - 4\int\frac{dx}{x^2+4}.$$ Then we can solve the second one of these integrals (first one is easy) by factoring out $4$ from the denominator and then using the substitution $u=\frac{x}{2}$: \begin{align*} \int\frac{dx}{x^2+4} & = \frac{1}{4}\int\frac{dx}{\left(\frac{x}{2}\right)^2+1}\\ &= \frac{1}{2}\int\frac{du}{u^2+1}.\end{align*} The last integral has a direct antiderivative.

For the second integral, try $x=2u$. Then $dx = 2du$, so $$\int\frac{1}{\sqrt{4-x^2}}\,dx = \int\frac{2du}{\sqrt{4-4u^2}} = \int\frac{du}{\sqrt{1-u^2}}.$$ Does the last one look familiar?

Alternatively, do the same trick: factor out $4$ in the denominator inside the square root: \begin{align*} \int\frac{1}{\sqrt{4-x^2}}\,dx &= \int\frac{1}{\sqrt{4(1 - (\frac{x}{2})^2}}\,dx\\ &= \int\frac{1}{2\sqrt{1 - (\frac{x}{2})^2}}\end{align*} which may suggest the substitution $w=\frac{x}{2}$.

My first impulse with the second integral would have been to use a trigonometric substitution. That's because integrals that involve $\sqrt{a^2-x^2}$ but which do not have an $x\,dx$ factor are the traditional proving grounds for the trigonometric substitution. Here, $a=2$, so we would try the substitution $x=2\sin\theta$, with $0\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, $dx = 2\cos\theta\,d\theta$. This wold turn the $\sqrt{4-x^2}$ factor into $$\sqrt{4-x^2} = \sqrt{4-4\sin^2\theta} = \sqrt{4(1-\sin^2\theta)} = \sqrt{4\cos^2\theta} = 2|\cos\theta|=2\cos\theta$$ with the last equality because $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, so $\cos\theta\geq 0$. Then the integral would simplify to $$\int\frac{dx}{\sqrt{4-x^2}} = \int\frac{\cos\theta\,d\theta}{2\cos\theta} = \frac{1}{2}\int\,d\theta,$$ which is trivial to do; finally, we would return to $x$ using $\theta=\frac{1}{2}\arcsin(x)$.

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Its better to use $x=2\sin u$ in the second integral – Tapu Jun 24 '12 at 1:58
thank you Arturo! and I also discovered about the \int already =) – Lucas Jun 24 '12 at 1:58
@Tapu: That's the standard trig substitution method; the answer will be exactly the same, but trigonometric substitution is usually not covered until after substutition. If these problems were substitution exercises, then presumably it is better to do them as intended. – Arturo Magidin Jun 24 '12 at 1:59
Also just another way showing that integrals can be evaluated in different methods quite often! – Joe Jun 24 '12 at 2:03
the teacher said that we could use all the methods that she taught us, and she did teach the method you used. I thought that trigonometric substitution was included in the 'substitution' definition, that's why I used the 'substitution' word. – Lucas Jun 24 '12 at 2:53

For the second one, use this substitution if you like:

$x=2 \sin u$. Then $dx=2\cos u\ du$.Substituting that into the second integral,we can reduce the problem like this :

$$\int \frac{1}{\sqrt{4-x^2}}dx=\int du=u+C$$[where C is the constant of integration]$$=\sin^{-1}\frac{x}{2}+C$$

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