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.
 A: Instead of applying another substitution, just simplify the expression after your first substitunion.
$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{10\sec^2{\theta}}{{\left(4\tan^2{\theta}+4\right)}^2} \; d \theta $$ $$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{10\sec^2{\theta}}{16{\left(\tan^2{\theta}+1\right)}^2} \; d \theta$$
$$=\frac{5}{8}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sec^2{\theta}}{{\sec^4{\theta}}} \; d \theta$$
$$=\frac{5}{8}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2{\theta} \; d \theta$$
Using the angle reduction formula for $\cos^2{\theta}$:
$$=\frac{5}{16} \left(\theta+\frac{1}{2}\sin{(2\theta)}\right) \bigg\rvert_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$$
$$=\frac{5\left(\pi+2\right)}{32}$$
A: 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
$$\int_{-2}^2 \dfrac{5}{(x^2+4)^2}dx$$$$=10\int_{0}^2 \dfrac{dx}{(x^2+2^2)^2}$$
$$=10\left[\frac{x}{2(2-1) 2^2(x^2+2^2)}+\frac{2\cdot 2-3}{2(2-1)2^2}\int \frac{dx}{x^2+2^2}\right]_0^2$$
$$=10\left[\frac{x}{8(x^2+4)}+\frac{1}{8}\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)\right]_0^2$$
$$=10\left[\frac{2}{64}+\frac{1}{16}\frac{\pi}{4}-0\right]$$
$$=\bbox[15px,#ffd,border:1px solid green]{\frac{5(\pi+2)}{32}}$$
A: Hint:
Remember that $\;1+\tan^2\theta=\dfrac1{\cos^2\theta}$, so you obtain
$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{10\sec^2θ}{(4\tan^2θ+4)^2}\,\mathrm dθ=\frac58 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sec^2θ}{(\tan^2θ+1)^2}\,\mathrm dθ =\frac58 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2θ\,\mathrm dθ,$$
and there remains to linearise the integrand with the duplication formulæ.
Added: The standard method to calculate the indefinite integral $I_n=\int\frac{\mathrm dx}{(a^2+x^2)^n}$ consists in establishing a recurrence relation between $I_n$ and $I_{n+1}$ by applying the inntegration by parts formula to the former integral.
A: Or, just integrate-by-parts,
\begin{align}
\int_{-2}^{2} \frac{5}{(x^2+4)^2}\,dx
&= \frac58\int_{-2}^{2} \frac1x d\left( \frac{x^2}{x^2+4} \right)\\
& = \frac58\frac x{x^2+4}\bigg|_{-2}^2+\frac54 \int_{-2}^{2} \frac1{x^2+4}dx
= \frac5{32}(2+\pi)
\end{align}
