<s>[INST] Definite integral of $\int_{-2}^{2} \frac{5}{(x^2+4)^2}\,dx$ using the substitution of $x=2\tanθ$. Can someone help with the integral $\int_{-2}^{2} \frac{5}{(x^2+4)^2}\,dx$?
I'm supposed to find the definite integral for this using the substitution $x=2\tanθ$.
This is what I've done so far:
$$\longrightarrow \frac{dx}{dθ}=2\sec^2θ$$
$$\longrightarrow x=2 \rightarrow θ=\frac{\pi}{4}$$
$$\longrightarrow x=-2 \rightarrow θ=-\frac{\pi}{4}$$
$$\therefore \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{10\sec^2θ}{(4\tan^2θ+4)^2}\,dθ$$
Using; $$t=\tan^2θ+1,$$
$$=\int_{2}^{2} \frac{10}{32t^2\sqrt{t-1}}\,dt$$
After this, I don't know how to finish. Does anyone know how to finish?
*Note: The first substitution is the one the exercise is telling me to use, the other is one I used myself.
 [/INST] Alternate easier method: use reduction formula:
$\color{blue}{\int \frac{dt}{(t^2+a^2)^n}=\frac{t}{2(n-1)a^2(t^2+a^2)^{n-1}}+\frac{2n-3}{2(n-1)a^2}\int \frac{dt}{(t^2+a^2)^{n-1}}}$   as follows
$$\int_{-2}^2 \dfrac{5}{(x^2+4)^2}dx$$$$=10\int_{0}^2 \dfrac{dx}{(x^2+2^2)^2}$$
$$=10\left[\frac{x}{2(2-1) 2^2(x^2+2^2)}+\frac{2\cdot 2-3}{2(2-1)2^2}\int \frac{dx}{x^2+2^2}\right]_0^2$$
$$=10\left[\frac{x}{8(x^2+4)}+\frac{1}{8}\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)\right]_0^2$$
$$=10\left[\frac{2}{64}+\frac{1}{16}\frac{\pi}{4}-0\right]$$
$$=\bbox[15px,#ffd,border:1px solid green]{\frac{5(\pi+2)}{32}}$$
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