The integral provided is:

$\int_0^\pi e^{\cos t} \sin(2t) \, dt$

We are asked to make a substitution and then use integration by parts to evaluate the integral. Let's follow these steps.

Step 1: Substitution

Consider the substitution:

$u = \cos t$ Then, the derivative is: $du = -\sin t \, dt$

From the original integral, we have $\sin(2t)$, which we can rewrite using a trigonometric identity:

$\sin(2t) = 2 \sin t \cos t$

So, the integral becomes:

$\int_0^\pi e^{\cos t} \cdot 2 \sin t \cos t \, dt$

Now, substitute $u = \cos t$, which implies $du = -\sin t \, dt$. The limits of integration also change:

• When $t = 0$, $u = \cos(0) = 1$.
• When $t = \pi$, $u = \cos(\pi) = -1$.

The integral now transforms to:

$-2 \int_1^{-1} e^u u \, du$

Step 2: Integration by Parts

We apply integration by parts to solve the integral. Recall the formula:

$\int u \, dv = uv - \int v \, du$

Let:

• $v = e^u$
• $du = u \, du$

Now, applying integration by parts:

The evaluated integral is:

$\int_0^\pi e^{\cos t} \sin(2t) \, dt = 4e^{-1}$

This means the final result of the integral is $4e^{-1}$, or approximately $4 \times 0.3679 = 1.4716$.

Would you like further details on the steps, or do you have any questions?

Here are 5 related questions for practice:

1. How would the result change if the limits of integration were from 0 to $\frac{\pi}{2}$?
2. What if the integrand was $e^{\sin t} \sin(2t)$?
3. Can you solve the integral using a different method besides integration by parts?
4. How does changing the power of the sine function in the integrand affect the result?
5. What is the result if you replace $\sin(2t)$ with $\cos(2t)$?

Tip: When using integration by parts, carefully choose the parts so that the resulting integrals simplify effectively.