Key Concepts and Formulas

• Properties of Definite Integrals: $\int_a^b f(x) dx + \int_a^b g(x) dx = \int_a^b (f(x) + g(x)) dx$.
• Trigonometric Identity: $1 + \tan^2 x = \sec^2 x$.
• Integration by Substitution: If $u = f(x)$, then $du = f'(x) dx$. The limits of integration must be changed accordingly.
• Limit of a Rational Function: $\mathop {\lim }\limits_{n \to \infty } \frac{P(n)}{Q(n)}$, where $P(n)$ and $Q(n)$ are polynomials in $n$.

Step-by-Step Solution

Step 1: Simplify the expression $I_n + I_{n+2}$ We are given $I_n = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx}$. We need to find a simplified form for $I_n + I_{n+2}$. $I_n + I_{n+2} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} + \int\limits_0^{\pi /4} {{{\tan }^{n + 2}}x\,dx}$ Using the property of definite integrals, we can combine these two integrals since they have the same limits of integration: $I_n + I_{n+2} = \int\limits_0^{\pi /4} \left( {{{\tan }^n}x + {{\tan }^{n + 2}}x} \right)\,dx$ Reasoning: The first step is to combine the given integral terms to simplify the expression that the limit will be applied to.

Step 2: Factor and apply trigonometric identity Inside the integral, we can factor out the common term $\tan^n x$: $I_n + I_{n+2} = \int\limits_0^{\pi /4} {{{\tan }^n}x \left( {1 + {{\tan }^2}x} \right)\,dx}$ Now, we use the fundamental trigonometric identity $1 + \tan^2 x = \sec^2 x$: $I_n + I_{n+2} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,{\sec ^2}x\,dx}$ Reasoning: Factoring out $\tan^n x$ is a strategic move to reveal a structure that can be simplified using a trigonometric identity. The presence of $\sec^2 x$, which is the derivative of $\tan x$, strongly suggests an integration by substitution.

Step 3: Evaluate the integral using substitution Let $u = \tan x$. Then, $du = \sec^2 x\,dx$. We need to change the limits of integration: When $x = 0$, $u = \tan(0) = 0$. When $x = \frac{\pi}{4}$, $u = \tan\left(\frac{\pi}{4}\right) = 1$. Substituting $u$ and $du$ into the integral, with the new limits: $I_n + I_{n+2} = \int\limits_0^1 {{u^n}\,du}$ Now, we evaluate this standard integral: $I_n + I_{n+2} = \left[ {\frac{{{u^{n + 1}}}}{{n + 1}}} \right]_0^1$ $I_n + I_{n+2} = \frac{{{{\left( 1 \right)}^{n + 1}}}}{{n + 1}} - \frac{{{{\left( 0 \right)}^{n + 1}}}}{{n + 1}}$ $I_n + I_{n+2} = \frac{1}{{n + 1}} - 0$ $I_n + I_{n+2} = \frac{1}{{n + 1}}$ Reasoning: The substitution $u = \tan x$ transforms the trigonometric integral into a simple power function integral, which is straightforward to evaluate using the power rule for integration. Crucially, the limits of integration are adjusted to match the new variable $u$.

Step 4: Evaluate the limit We need to find the limit of $n\left[ {{I_n} + {I_{n + 2}}} \right]$ as $n \to \infty$. Substitute the simplified expression for $I_n + I_{n+2}$: $\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right] = \mathop {\lim }\limits_{n \to \infty } \,n\left[ {\frac{1}{{n + 1}}} \right]$ $= \mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + 1}}$ To evaluate this limit, we divide the numerator and the denominator by the highest power of $n$ in the denominator, which is $n$: $= \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{n}{n}}}{{\frac{n}{n} + \frac{1}{n}}}$ $= \mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 + \frac{1}{n}}}$ As $n \to \infty$, the term $\frac{1}{n} \to 0$. $= \frac{1}{{1 + 0}}$ $= 1$ Reasoning: This step involves evaluating a limit of a rational function as $n$ approaches infinity. The standard technique is to divide all terms by the highest power of $n$ in the denominator, which simplifies the expression and allows for the direct evaluation of the limit.

Common Mistakes & Tips

• Forgetting to change limits of integration: When performing substitution in definite integrals, always update the limits of integration to correspond to the new variable.
• Incorrectly applying trigonometric identities: Ensure you are using the correct identities, like $1 + \tan^2 x = \sec^2 x$.
• Algebraic errors in limit calculation: Be meticulous when dividing by the highest power of $n$ and simplifying the rational expression for the limit.

Summary

The problem requires evaluating the limit of $n$ times the sum of two consecutive integrals $I_n$ and $I_{n+2}$. The key to solving this problem lies in first simplifying the sum $I_n + I_{n+2}$. By combining the integrals and factoring, we obtain $\int\limits_0^{\pi /4} {{{\tan }^n}x\,{\sec ^2}x\,dx}$. This integral is readily solved using the substitution $u = \tan x$. After evaluating the integral to be $\frac{1}{n+1}$, we then compute the limit $\mathop {\lim }\limits_{n \to \infty } n \left(\frac{1}{n+1}\right)$, which simplifies to 1.

The final answer is $\boxed{1}$.