Evaluate the integral.
-$$\int _ { 0 } ^ { \pi / 2 } 3 ^ { \cos t } \sin t d t$$
A) $\frac { - 2 } { \ln 3 }$
B) $\frac { 3 ^ { \pi / 2 } - 1 } { \ln 3 }$
C) $\frac { 2 } { \ln 3 }$
D) 2

Question

Quizplus · Accepted Answer

To solve the integral $\int _ { 0 } ^ { \pi / 2 } 3 ^ { \cos t } \sin t d t$, we can use a substitution method. Let $u = \cos(t)$, which implies $du = -\sin(t) dt$. The limits of integration change accordingly: when $t = 0$, $u = 1$, and when $t = \pi/2$, $u = 0$. Thus, the integral becomes $-\int _ { 1 } ^ { 0 } 3 ^ { u } du$, which is equivalent to $\int _ { 0 } ^ { 1 } 3 ^ { u } du$. This integral can be evaluated as $\frac{3^u}{\ln(3)}\Big|_0^1 = \frac{3 - 1}{\ln(3)} = \frac{2}{\ln(3)}$, making option C the correct answer.

Evaluate the Integral ∫0π/23cos⁡tsin⁡tdt\int _ { 0 } ^ { \pi / 2 } 3 ^ { \cos t } \sin t d t∫0π/2​3costsintdt

Evaluate the Integral $\int _ { 0 } ^ { \pi / 2 } 3 ^ { \cos t } \sin t d t$