The expression in the image is:

$\int \frac{\sin x \cdot e^{\sin x} - 1}{\sin^2 x} \, dx$

To solve this integral, we can attempt a substitution. Let's define $u = \sin x$. Therefore, $du = \cos x \, dx$.

The integral becomes:

$\int \frac{u \cdot e^{u} - 1}{u^2} \cdot \frac{dx}{du} \, du$

However, since $\frac{dx}{du}$ depends on the specific relationship between $x$ and $u$ (which involves $\cos x$), the substitution does not simplify easily without further steps.

Instead, we can simplify the integral by splitting it:

$\int \frac{\sin x \cdot e^{\sin x}}{\sin^2 x} \, dx - \int \frac{1}{\sin^2 x} \, dx$

The second integral can be rewritten using the identity $\csc^2 x$ (since $\frac{1}{\sin^2 x} = \csc^2 x$):

$\int \csc^2 x \, dx = -\cot x$

For the first integral, notice that it resembles the derivative of $\frac{-e^{\sin x}}{\sin x}$. Hence, let's confirm:

$\frac{d}{dx} \left(-\frac{e^{\sin x}}{\sin x}\right) = \frac{\sin x \cdot e^{\sin x} - e^{\sin x} \cdot \cos x}{\sin^2 x} = \frac{\sin x \cdot e^{\sin x} - 1}{\sin^2 x}$

This shows that the first integral is:

$\int \frac{\sin x \cdot e^{\sin x}}{\sin^2 x} \, dx = -\frac{e^{\sin x}}{\sin x} + C_1$

Thus, the full solution is:

$-\frac{e^{\sin x}}{\sin x} + \cot x + C$

Do you want more details on any part of this solution?

Related Questions:

1. How does the substitution method simplify complex integrals?
2. What other methods can be used to solve similar integrals?
3. What is the significance of trigonometric identities in solving integrals?
4. How can we handle integrals involving exponential functions more effectively?
5. What are some other common integrals involving $\csc^2 x$?

Tip: When dealing with integrals involving trigonometric functions, always look for identities or substitution methods that can simplify the expression.