Calculus - Integral cosx e^ sinx and Integral sinx e^ cosx

The integrals of functions involving trigonometric and exponential expressions can be quite intricate and fascinating. Let's delve into the integrals of \( \cos(x) e^{\sin(x)} \) and \( \sin(x) e^{\cos(x)} \).

1. **Integral of \( \cos(x) e^{\sin(x)} \)**:
The integral of \( \cos(x) e^{\sin(x)} \) can be approached using a substitution method. Let \( u = \sin(x) \). Then, \( du = \cos(x) dx \). This transforms the integral into a more straightforward form:
\[ \int \cos(x) e^{\sin(x)} dx = \int e^u du. \]
The integral of \( e^u \) is simply \( e^u \). Therefore, substituting back \( u = \sin(x) \), we get:
\[ \int \cos(x) e^{\sin(x)} dx = e^{\sin(x)} + C, \]
where \( C \) is the constant of integration.

2. **Integral of \( \sin(x) e^{\cos(x)} \)**:
Similarly, for the integral of \( \sin(x) e^{\cos(x)} \), we also use a substitution approach. Let \( u = \cos(x) \). Then, \( du = -\sin(x) dx \), which means \( -du = \sin(x) dx \). Substitute these into the integral:
\[ \int \sin(x) e^{\cos(x)} dx = \int -e^u du. \]
The integral of \( -e^u \) is \( -e^u \). Substituting back \( u = \cos(x) \), we obtain:
\[ \int \sin(x) e^{\cos(x)} dx = -e^{\cos(x)} + C, \]
where \( C \) is the constant of integration.

In summary, the integrals of these functions leverage the power of substitution to simplify the integration process, yielding elegant results that combine trigonometric and exponential functions.