# How to evaluete: int_0^(pi/2) e^(sinx)*cosxdx ?

Oct 6, 2017

Use substitution to get the answer $e - 1 \approx 1.718$.

#### Explanation:

One way to do this is to first find the indefinite integral before applying the Fundamental Theorem of Calculus. The indefinite integral can be found by substituting $u = \sin \left(x\right) , \mathrm{du} = \cos \left(x\right) \mathrm{dx}$ to get

$\int \setminus {e}^{\sin \left(x\right)} \cdot \cos \left(x\right) \setminus \mathrm{dx} = \int \setminus {e}^{u} \setminus \mathrm{du} = {e}^{u} + C = {e}^{\sin \left(x\right)} + C$

Therefore, ${\int}_{0}^{\frac{\pi}{2}} {e}^{\sin \left(x\right)} \cdot \cos \left(x\right) \setminus \mathrm{dx} = {e}^{\sin \left(\frac{\pi}{2}\right)} - {e}^{\sin \left(0\right)} = {e}^{1} - 1 = e - 1 \approx 1.718$.

Alternatively, you could keep it as a definite integral and change the limits of integration after using the same substitution $u = \sin \left(x\right) , \mathrm{du} = \cos \left(x\right) \mathrm{dx}$ as follows:

${\int}_{0}^{\frac{\pi}{2}} {e}^{\sin \left(x\right)} \cdot \cos \left(x\right) \setminus \mathrm{dx} = {\int}_{0}^{1} {e}^{u} \setminus \mathrm{du} = {e}^{1} - {e}^{0} = e - 1 \approx 1.718$.

Oct 6, 2017

$e - 1.$

#### Explanation:

Let, $I = {\int}_{0}^{\frac{\pi}{2}} {e}^{\sin x} \cdot \cos x \mathrm{dx} .$

Subst., $\sin x = t \Rightarrow \cos x \mathrm{dx} = \mathrm{dt} .$

Also, $x = 0 \Rightarrow t = \sin 0 = 0 , \mathmr{and} , x = \frac{\pi}{2} \Rightarrow t = \sin \left(\frac{\pi}{2}\right) = 1.$

$\therefore I = {\int}_{0}^{1} {e}^{t} \mathrm{dt} ,$

$= {\left[{e}^{t}\right]}_{0}^{1} ,$

$= {e}^{1} - {e}^{0.}$

$\Rightarrow I = e - 1.$