Key Concepts and Formulas

• Substitution Method for Definite Integrals: This technique is used to simplify integrals by changing the variable of integration. It is essential to update the limits of integration according to the substitution.
• Properties of Definite Integrals: Specifically, the property $\int_a^b f(x) dx = -\int_b^a f(x) dx$ is used to reverse the limits of integration, and the property $\int_{-a}^a f(x) dx = \int_{-a}^a f(-x) dx$ is useful for integrals with symmetric limits.
• Algebraic Simplification: Basic algebraic manipulation of exponents and fractions is required.

Step-by-Step Solution

Step 1: Analyze the Integral and Apply Substitution

The given integral is: $I = \int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx}$ We observe that the numerator contains $\sin x \, dx$, and the terms in the integrand are functions of $\cos x$. This suggests using the substitution $t = \cos x$.

Let $t = \cos x$. Differentiating both sides with respect to $x$, we get $\frac{dt}{dx} = -\sin x$, which implies $dt = -\sin x \, dx$, or $\sin x \, dx = -dt$.

Now, we must change the limits of integration: When $x = 0$, $t = \cos(0) = 1$. When $x = \pi$, $t = \cos(\pi) = -1$.

Substituting these into the integral, we get: $I = \int\limits_1^{-1} {{{{e^t}(-dt)} \over {(1 + {t^2})({e^t} + {e^{ - t}})}}}$ $I = - \int\limits_1^{-1} {{{{e^t}} \over {(1 + {t^2})({e^t} + {e^{ - t}})}}dt}$ Using the property $\int_a^b f(x) dx = -\int_b^a f(x) dx$, we can swap the limits of integration and remove the negative sign: $I = \int\limits_{-1}^1 {{{{e^t}} \over {(1 + {t^2})({e^t} + {e^{ - t}})}}dt} \quad \text{... (Equation 1)}$

Step 2: Apply the Property of Definite Integrals for Symmetric Limits

The integral in Equation 1 has symmetric limits from $-1$ to $1$. We can use the property $\int_{-a}^a f(t) \, dt = \int_{-a}^a f(-t) \, dt$. Let $f(t) = {{{e^t}} \over {(1 + {t^2})({e^t} + {e^{ - t}})}}$. Then, $f(-t) = {{{e^{-t}}} \over {(1 + {(-t)^2})({e^{-t}} + {e^{ - (-t)}})}} = {{{e^{-t}}} \over {(1 + {t^2})({e^{-t}} + {e^t})}}$.

Therefore, we can also write the integral $I$ as: $I = \int\limits_{-1}^1 {{{{e^{-t}}} \over {(1 + {t^2})({e^{-t}} + {e^t})}}dt} \quad \text{... (Equation 2)}$

Step 3: Combine the Two Forms of the Integral

Now, we add Equation 1 and Equation 2: $I + I = \int\limits_{-1}^1 {{{{e^t}} \over {(1 + {t^2})({e^t} + {e^{ - t}})}}dt} + \int\limits_{-1}^1 {{{{e^{-t}}} \over {(1 + {t^2})({e^{-t}} + {e^t})}}dt}$ $2I = \int\limits_{-1}^1 {\left( {{{e^t}} \over {(1 + {t^2})({e^t} + {e^{ - t}})}} + {{{e^{-t}}} \over {(1 + {t^2})({e^{-t}} + {e^t})}} \right)} dt$ Since the denominators are the same, we can combine the numerators: $2I = \int\limits_{-1}^1 {{{{e^t} + {e^{-t}}} \over {(1 + {t^2})({e^t} + {e^{ - t}})}} dt}$ Notice that the term $(e^t + e^{-t})$ appears in both the numerator and the denominator. We can cancel this term, provided $e^t + e^{-t} \neq 0$, which is always true for real $t$. $2I = \int\limits_{-1}^1 {{{{1}} \over {(1 + {t^2})}} dt}$

Step 4: Evaluate the Simplified Integral

The integral is now a standard form: $2I = \int\limits_{-1}^1 \frac{1}{1+t^2} dt$ The antiderivative of $\frac{1}{1+t^2}$ is $\tan^{-1}(t)$. $2I = \left[ \tan^{-1}(t) \right]_{-1}^1$ Now, we evaluate the definite integral using the limits: $2I = \tan^{-1}(1) - \tan^{-1}(-1)$ We know that $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(-1) = -\frac{\pi}{4}$. $2I = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right)$ $2I = \frac{\pi}{4} + \frac{\pi}{4}$ $2I = \frac{2\pi}{4}$ $2I = \frac{\pi}{2}$

Step 5: Solve for I

Finally, we solve for $I$ by dividing by 2: $I = \frac{1}{2} \times \frac{\pi}{2}$ $I = \frac{\pi}{4}$

Common Mistakes & Tips

• Forgetting to change limits: When using substitution in definite integrals, always update the limits of integration to correspond to the new variable.
• Sign errors with substitution: The derivative of $\cos x$ is $-\sin x$, so remember to include the negative sign when performing the substitution $dt = -\sin x \, dx$.
• Misapplying the symmetric limit property: Ensure the integrand is indeed an even function of the variable $t$ when using $\int_{-a}^a f(t) dt = \int_{-a}^a f(-t) dt$ or an odd function for $\int_{-a}^a f(t) dt = 0$. In this case, we used the property to get an alternative form of the integral and then added them.

Summary

The problem is solved by first applying a trigonometric substitution to simplify the integrand and transform the limits of integration. The resulting integral has symmetric limits, allowing us to use a property of definite integrals to obtain a second form of the integral. By adding the original and transformed integrals, the expression simplifies significantly. The final step involves evaluating a standard integral of $\frac{1}{1+t^2}$ and solving for the value of the original integral.

The final answer is $\boxed{{\frac{\pi}{4}}}$.