I was just calculating an integral via a trigonometric substitution and ended up with $\color{red}{ \text{something pretty nonsensical} }$ but $\color{blue}{ \text{reversing the substitution} }$ seemed to clean it up.
$$\begin{aligned} \int_{0}^{\frac{\pi}{2}} \dfrac{\text{d}\theta}{3+5\cos \theta} \ & \overset{t=\tan \frac{\theta}{2}}= \ \dfrac{1}{4} \color{red}{ \log \left| \dfrac{2+t}{2-t} \right| \Bigg|}_{\color{red}{0}}^{ \color{purple}{\infty }} \\ & \ \ = \dfrac{1}{4} \color{blue}{ \log \left| \dfrac{2+\tan\frac{\theta}{2}}{2-\tan\frac{\theta}{2}} \right| \Bigg|_{0}^{\frac{\pi}{2}} } \end{aligned}$$
Why is that the case? Is it something to do with the nature of the substitution?
Is there something I'm not considering when performing the substitution?
Any help would be much appreciated.
Thank you.
Edit: It turns out that $\color{purple}{\tan\frac{\pi}{4} =1}$. Problem solved!
Answer
When $\theta$ goes from $0$ to $\frac{\pi}2$, $\frac{\theta}2$ goes from $0$ to $\frac{\pi}4$ so $\tan\frac{\theta}2$ goes from $0$ t0 $1$, not $0$ to $\infty$.
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