Markus wrote:$(1)$ $$ \int \frac{\sin x + 2\cos x}{3\sin x + 4\cos x} \, \text{d}x$$
$(2) \enspace \text{(Putnam)}$ $$\int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}} \, \text{d}x$$
Gjerne kom med oppfølgere!
$(1)$
Bruker først substitusjonen $\tan \left(\frac{x}{2}\right) = t$, deretter brøkoppspalting: $$\int \frac{\sin x + 2\cos x}{3\sin x + 4\cos x} \, \text{d}x = \int\frac{\frac{2t}{1+t^2} + 2\frac{1-t^2}{1+t^2}}{3\frac{2t}{1+t^2} +4\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}\, \text{d}t = 2\int\frac{t^2-t-1}{\left(1+t^2\right)\left(2t^2-3t-2\right)}\, \text{d}t = 2\int\frac{t^2-t-1}{\left(1+t^2\right)\left(2t+1\right)\left(t-2\right)}\, \text{d}t \\ = 2\int\left[\frac{11-2t}{25(1+t^2)} + \frac{1}{25(t-2)} + \frac{2}{25(2t+1)}\right]\, \text{d}t = \frac{22}{25}\arctan(t) - \frac{4}{25}\int\frac{t}{1+t^2}\, \text{d}t + \frac{2}{25}\log(t-2) + \frac{2}{25}\log(2t+1) + C \\ = \frac{11}{25}x -\frac{2}{25}\log(t^2+1) + \frac{2}{25}\log(t-2) + \frac{2}{25}\log(2t+1) + C = \frac{1}{25}\left[11x + 2\log\left(\frac{(t-2)(2t+1)}{1+t^2}\right)\right]+C \\ = \frac{1}{25}\left[11x + 2\log\left(3\sin x + 4\cos x\right)\right]+C \\ $$
$(2)$ Utnytter antisymmetrien til integranden om $x=3$ og substituerer $u=x-3$:
$$\begin{align*} \int_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}} \, \text{d}x & = \int_{-1}^1\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}} \, \text{d}u \\
& = \int_{-1}^0\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}} \, \text{d}u + \int_0^1\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}} \, \text{d}u \\
& = \int_0^1\frac{\sqrt{\ln(6+v)}}{\sqrt{\ln(6+v)}+\sqrt{\ln(6-v)}} \, \text{d}v + \int_0^1\frac{\sqrt{\ln(6-u)}}{\sqrt{\ln(6-u)}+\sqrt{\ln(6+u)}} \, \text{d}u \text{ }\text{ }\text{ (etter substitusjonen }v=-u\text{)} \\
& = \int_0^1\frac{\sqrt{\ln(6+x)} + \sqrt{\ln(6-x)}}{\sqrt{\ln(6-x)} + \sqrt{\ln(6+x)}}\, \text{d}x \\
& = \int_0^1\, \text{d}x \\
& = 1.
\end{align*}$$
