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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}


Learn more about this topic:

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Integration Problems in Calculus: Solutions & Examples

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Chapter 13 / Lesson 13
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Learn what integration problems are. Discover how to find integration sums and how to solve integral calculus problems using calculus example problems.


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