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.
Don't know if you really meant
∫sin(t)e^t dt
If that's the case, try differentiate using the chain rule,
sin(t)e^t
and
cos(t)e^t
You should be able to figure out the integral from the results.
The original integral you posted does not seem to have an analytic solution. A series solution is (almost always) available but does not fit your bill.
uh no i meant what i typed which is e^(sin(t))
In that case, I fear there is no solution using elementary functions.
who posed such a problem? It's solvable numerically, using a Taylor series, but not symbolically.
To solve this problem, you don't necessarily need to use the chain rule. Instead, you can directly apply the fundamental theorem of calculus.
The fundamental theorem of calculus states that if F(x) is an antiderivative of a function f(x), then ∫[a to b] f(x) dx = F(b) - F(a).
In this case, we have z = ∫[x to y] e^sin(t) dt.
To find an antiderivative of e^sin(t), you can use the technique of integration by substitution.
Let u = sin(t), then du/dt = cos(t), and dt = du/cos(t).
Substituting these values into the integral, we have:
z = ∫[x to y] e^sin(t) dt
= ∫[x to y] e^u (du/cos(t))
= ∫[x to y] e^u du/cos(t)
Since we need to find the integral with respect to t, we can replace cos(t) with √(1 - sin^2(t)) using the Pythagorean identity for cosine.
Now, we have:
z = ∫[x to y] e^u du/√(1 - u^2)
To evaluate this integral, you need to find the antiderivative of e^u/√(1 - u^2) with respect to u. Let's call this antiderivative G(u).
Now, applying the fundamental theorem of calculus, we have:
z = G(u) ∣ [x to y]
= G(y) - G(x)
So, to get the value of z, you need to find the antiderivative G(u), evaluate it at y and x, and then subtract the two values.