# Evaluate the following function. {eq}\displaystyle \int \dfrac {e^{\cos x} \sin x}{1 + e^{\cos x}} dx {/eq}

## Question:

Evaluate the following function.

{eq}\displaystyle \int \dfrac {e^{\cos x} \sin x}{1 + e^{\cos x}} dx {/eq}

## Indefinite Integral:

The indefinite integral is the inverse operation of differentiation. The indefinite integral finds a function whose derivative is the function being integrated. Normally, in a definite integral, the constants of a function cannot be determined, and thus, a generic constant of integration is included.

## Answer and Explanation: 1

Given:

{eq}\displaystyle \int\ \frac{e^{\cos\ x} \sin\ x}{1 + e^{\cos\ x}}\ dx {/eq}

Let:

{eq}\displaystyle u = 1 + e^{\cos\ x} {/eq}

The derivative of this variable is:

{eq}\displaystyle du = e^{\cos\ x} (-\sin\ x)\ dx {/eq}

We can rewrite the integral in terms of u:

{eq}\displaystyle \int\ \frac{e^{\cos\ x} \sin\ x}{1 + e^{\cos\ x}}\ dx = -\int\ \frac{1}{u}\ du {/eq}

This is a standard integral:

{eq}\displaystyle \int\ \frac{1}{u}\ du = \ln |u| + C {/eq}

So, we get;

{eq}\displaystyle \int\ \frac{e^{\cos\ x} \sin\ x}{1 + e^{\cos\ x}}\ dx = -\ln |u| + C {/eq}

Return the definition of u to get:

{eq}\displaystyle \boxed{\int\ \frac{e^{\cos\ x} \sin\ x}{1 + e^{\cos\ x}}\ dx = -\ln |1+ e^{\cos\ x} | + C} {/eq}