Evaluate the integral.
-$\int _ { 0 } ^ { \pi / 2 } 5 ^ { \cos t } \sin t d t$
A) $\frac { 5 ^ { \pi / 2 } - 1 } { \ln 5 }$
B) 4
C) $\frac { 4 } { \ln 5 }$
D) $\frac { - 4 } { \ln 5 }$

Question

Quizplus · Accepted Answer

We can use substitution $u=cos(t)$ and $du=-sin(t)dt$ which gives us $\int_{1}^{0}-5^{u}du$. Now, we can use the formula for the geometric series:
$$\sum_{n=0}^{\infty}cr^{n}= \frac{c}{1-r}$$
to get the integral value:
\begin{align*} \int_{0}^{\frac{\pi}{2}} 5^{\cos t} \sin t dt &=\int_{1}^{0}-5^{u}du \ &=\frac{1}{\ln 5}\int_{1}^{0} -d(5^{u}) \ &=\frac{1}{\ln 5}[5^{0}-5^{1}] \ &=\boxed{\frac{4}{\ln 5}} \end{align*}

Evaluate the Integral ∫0π/25cos⁡tsin⁡tdt\int _ { 0 } ^ { \pi / 2 } 5 ^ { \cos t } \sin t d t∫0π/2​5costsintdt

Evaluate the Integral $\int _ { 0 } ^ { \pi / 2 } 5 ^ { \cos t } \sin t d t$