How do you evaluate the definite integral #int cosxe^sinx# from #[0, pi/2]#?

Question

Emma Aldridge · Accepted Answer

#e-1# #int_o^(pi/2)cosxe^sinxdx# recognising that # d/(dx)(e^sinx)=cosxe^sinx # we have #int_o^(pi/2)cosxe^sinxdx=[e^sinx]_0^(pi/2)# #=e^sin(pi/2)-e^sin0# #=e^1-e^0# #=e-1#

Cameron Alexander · Answer

undefined To evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos(x)e^{\sin(x)} \,dx$, you can use the technique of integration by parts. Let $u = \cos(x)$ and $dv = e^{\sin(x)}\,dx$. Then, $du = -\sin(x)\,dx$ and $v = -e^{\sin(x)}$.

Now, applying the integration by parts formula, $\int u \,dv = uv - \int v \,du$, we get:

$$
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \cos(x)e^{\sin(x)} \,dx &= \left. -\cos(x)e^{\sin(x)} \right|_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} (-e^{\sin(x)})\sin(x) \,dx \
&= -\cos\left(\frac{\pi}{2}\right)e^{\sin\left(\frac{\pi}{2}\right)} + \cos(0)e^{\sin(0)} - \int_{0}^{\frac{\pi}{2}} (-e^{\sin(x)})\sin(x) \,dx \
&= 0 - e^0 - \int_{0}^{\frac{\pi}{2}} (-e^{\sin(x)})\sin(x) \,dx \
&= -1 - \int_{0}^{\frac{\pi}{2}} (-e^{\sin(x)})\sin(x) \,dx
\end{align*}
$$

Now, for the remaining integral, $\int_{0}^{\frac{\pi}{2}} (-e^{\sin(x)})\sin(x) \,dx$, there's no straightforward method to evaluate it analytically. Thus, we might resort to numerical methods like Simpson's rule or the trapezoidal rule to approximate its value.