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

2 Answers
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(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.

Oct 6, 2017

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.