Question

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

Answer 1

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(x), du=cos(x)dx` to get

`int\ e^{sin(x)}*cos(x)\ dx=int\ e^u\ du=e^u+C=e^{sin(x)}+C`

Therefore, `int_{0}^{pi/2}e^{sin(x)}*cos(x)\ dx=e^{sin(pi/2)}-e^{sin(0)}=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(x), du=cos(x)dx` as follows:

`int_{0}^{pi/2}e^{sin(x)}*cos(x)\ dx=int_{0}^{1}e^{u}\ du=e^{1}-e^{0}=e-1 approx 1.718`.

Answer 2

`e-1.`

Explanation:

Let, `I=int_0^(pi/2)e^(sinx)*cosxdx.`

Subst., `sinx=t rArr cosxdx=dt.`

Also, `x=0 rArr t=sin0=0, and, x=pi/2 rArr t=sin(pi/2)=1.`

`:. I=int_0^1e^tdt,`

`=[e^t]_0^1,`

`=e^1-e^0.`

`rArr I=e-1.`