Use fundamental theorem of calculus:
Int(pi/2_pi) of e^(sin(q))*cos(q)dq
u-substitute u=sin(q). du = cos(q) dq.
from u=sin(pi)=0 to u=sin(2pi)=0
int(e^u du) from 0 to 0
=e^0-e^0
=0
To evaluate the integral ∫(π/2 to π) e^(sin(q))*cos(q) dq using the fundamental theorem of calculus, we first need to find the antiderivative of the function e^(sin(q))*cos(q) with respect to q. This antiderivative will allow us to evaluate the definite integral between the given limits.
Let F(q) be the antiderivative of e^(sin(q))*cos(q). By applying integration techniques and the chain rule, we can find F(q) as follows:
1. Notice that the given function, e^(sin(q))*cos(q), is in the form of the product of two functions, where one is the derivative of the other. This suggests the use of u-substitution.
Let u = sin(q), then du = cos(q) dq.
2. Rewrite the integral in terms of the new variable u:
∫e^u du
3. Integrate e^u with respect to u:
F(q) = ∫e^u du = e^u + C
where C is the constant of integration.
4. Substitute back sin(q) for u:
F(q) = e^(sin(q)) + C
Now that we have the antiderivative F(q) = e^(sin(q)) + C, we can evaluate the definite integral using the fundamental theorem of calculus.
∫(π/2 to π) e^(sin(q))*cos(q) dq = F(π) - F(π/2)
= [e^(sin(π)) + C] - [e^(sin(π/2)) + C]
= [e^(0) + C] - [e^(1) + C]
= 1 - e + C - C
= 1 - e
Therefore, using the fundamental theorem of calculus, the value of the given integral is 1 - e.