Answer
$$\int^{\pi/2}_{0}7^{\cos t}\sin tdt=\frac{6}{\ln7}$$
Work Step by Step
$$A=\int^{\pi/2}_{0}7^{\cos t}\sin tdt$$
We set $\cos t=u$, which means $$-\sin tdt=du$$ $$\sin tdt=-du$$
- For $t=\pi/2$, we have $u=\cos(\pi/2)=0$
- For $t=0$, we have $u=\cos0=1$
Therefore, $$A=-\int^{0}_{1}7^udu$$ $$A=-\frac{7^u}{\ln7}\Big]^{0}_{1}$$ $$A=-\frac{1}{\ln7}(7^0-7^1)$$ $$A=-\frac{-6}{\ln7}=\frac{6}{\ln7}$$
