<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] 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 $\cos^2{\theta}$:
$$=\frac{5}{16} \left(\theta+\frac{1}{2}\sin{(2\theta)}\right) \bigg\rvert_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$$
$$=\frac{5\left(\pi+2\right)}{32}$$
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