Let z = ∫e^(sin(t))dt from x to y
a = x
b = y
I tried thinking about it like a chain rule but even then i'm a little unsure.
I know dz/dt = e^(sin(t)). Can you please point me in the right direction if i'm supposed to use the chain rule.
To evaluate the integral ∫e^(sin(t))dt from x to y, you can indeed apply the chain rule. However, in this case, you would need to use the chain rule in reverse, which is called u-substitution. Let me explain how to do it step by step:
1. Let u = sin(t). Taking the derivative of u with respect to t gives du/dt = cos(t). Rearrange this expression to solve for dt: dt = du/cos(t).
2. Now substitute the value of dt in the integral. Since u = sin(t), we have du = cos(t) dt. Substitute du = cos(t) dt into the integral to get ∫e^u * (du/cos(t)). Notice that the cos(t) in the denominator cancels out with the one in the numerator.
3. Simplify the integral to ∫e^u du. This is now a much simpler integral to evaluate.
4. Integrate ∫e^u du with respect to u. The antiderivative of e^u is simply e^u. So, the integral becomes e^u + C, where C is the constant of integration.
5. Finally, substitute back for u. Recall that u = sin(t). So the final result is: e^sin(t) + C.
To find the definite integral from x to y, you can evaluate the antiderivative at the upper bound (y) and subtract the value of the antiderivative at the lower bound (x). So, the value of the definite integral is: e^sin(y) - e^sin(x).