Answer
$$\int^{\pi/4}_{0}\Big(\frac{1}{3}\Big)^{\tan t}\sec^2 tdt=\frac{2}{3\ln3}$$
Work Step by Step
$$A=\int^{\pi/4}_{0}\Big(\frac{1}{3}\Big)^{\tan t}\sec^2 tdt$$
We set $\tan t=u$, which means $$\sec^2 tdt=du$$
- For $t=\pi/4$, we have $u=\tan(\pi/4)=1$
- For $t=0$, we have $u=\tan0=0$
Therefore, $$A=\int^{1}_{0}\Big(\frac{1}{3}\Big)^udu$$ $$A=\frac{\Big(\frac{1}{3}\Big)^u}{\ln\frac{1}{3}}\Big]^{1}_{0}$$ $$A=\frac{1}{\ln\frac{1}{3}}(\frac{1}{3}-1)$$ $$A=-\frac{2}{3\ln\frac{1}{3}}$$
We can further rewrite $\ln(1/3)=\ln1-\ln3=-\ln3$
Thus, $$A=\frac{2}{3\ln3}$$
