Key Concepts and Formulas

• Integration by Parts: The formula for definite integrals is $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$.
• Trigonometric Identity: $\sec^2 x = 1 + \tan^2 x$.
• Recurrence Relation: A relation that defines a sequence where each term is a function of preceding terms. This is often derived using integration by parts.

Step-by-Step Solution

Step 1: Define a General Integral and Rewrite the Expression To simplify the given expression, let's define a general integral $I(n)$ that captures the common structure. The given expression is: $\frac{e^{-\pi/4} + \int_0^{\pi/4} e^{-x} \tan^{50} x \, dx}{\int_0^{\pi/4} e^{-x} (\tan^{49} x + \tan^{51} x) \, dx}$ We define $I(n) = \int_0^{\pi/4} e^{-x} \tan^n x \, dx$. Using this notation, the expression becomes: $\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)}$ Our goal is to find a relationship between $I(n)$, $I(n-1)$, and $I(n+1)$.

Step 2: Apply Integration by Parts to $I(n)$ We apply integration by parts to $I(n) = \int_0^{\pi/4} e^{-x} \tan^n x \, dx$. We choose:

• $u = \tan^n x$ (because its derivative simplifies the power)
• $dv = e^{-x} \, dx$ (because it's easily integrable)

Then, $du = n \tan^{n-1} x \sec^2 x \, dx$ and $v = \int e^{-x} \, dx = -e^{-x}$. Applying the integration by parts formula: $I(n) = \left[ -e^{-x} \tan^n x \right]_0^{\pi/4} - \int_0^{\pi/4} (-e^{-x}) (n \tan^{n-1} x \sec^2 x) \, dx$ $I(n) = \left[ -e^{-x} \tan^n x \right]_0^{\pi/4} + n \int_0^{\pi/4} e^{-x} \tan^{n-1} x \sec^2 x \, dx$

Step 3: Evaluate the Definite Part and Simplify the Integral Term Evaluate the definite part $\left[ -e^{-x} \tan^n x \right]_0^{\pi/4}$: At $x = \pi/4$: $-e^{-\pi/4} \tan^n(\pi/4) = -e^{-\pi/4} (1)^n = -e^{-\pi/4}$. At $x = 0$: $-e^{-0} \tan^n(0) = -1 \cdot 0^n = 0$ (since $n=50 > 0$). So, the definite part is $-e^{-\pi/4} - 0 = -e^{-\pi/4}$.

Now, simplify the integral term using the identity $\sec^2 x = 1 + \tan^2 x$: $n \int_0^{\pi/4} e^{-x} \tan^{n-1} x (1 + \tan^2 x) \, dx$ $n \int_0^{\pi/4} e^{-x} (\tan^{n-1} x + \tan^{n-1} x \tan^2 x) \, dx$ $n \int_0^{\pi/4} e^{-x} (\tan^{n-1} x + \tan^{n+1} x) \, dx$ We can split this into two integrals: $n \left( \int_0^{\pi/4} e^{-x} \tan^{n-1} x \, dx + \int_0^{\pi/4} e^{-x} \tan^{n+1} x \, dx \right)$ Using our definition $I(k) = \int_0^{\pi/4} e^{-x} \tan^k x \, dx$, this becomes: $n (I(n-1) + I(n+1))$

Step 4: Formulate the Recurrence Relation Combining the results from Step 2 and Step 3, we get the recurrence relation: $I(n) = -e^{-\pi/4} + n(I(n-1) + I(n+1))$

Step 5: Substitute the Recurrence Relation into the Original Expression We need to evaluate the expression $\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)}$. Let's substitute $n=50$ into our recurrence relation: $I(50) = -e^{-\pi/4} + 50(I(49) + I(51))$ Now, substitute this expression for $I(50)$ into the numerator of the given expression: $\frac{e^{-\pi/4} + \left( -e^{-\pi/4} + 50(I(49) + I(51)) \right)}{I(49) + I(51)}$ The $e^{-\pi/4}$ terms in the numerator cancel out: $\frac{50(I(49) + I(51))}{I(49) + I(51)}$ Since $e^{-x} > 0$ and $\tan x > 0$ for $x \in (0, \pi/4)$, the integral $I(k) = \int_0^{\pi/4} e^{-x} \tan^k x \, dx$ is positive for any $k \ge 0$. Therefore, $I(49) + I(51) \neq 0$. We can cancel out the term $(I(49) + I(51))$ from the numerator and denominator: $= 50$

The value of the given expression is 50.

Common Mistakes & Tips

• Incorrect choice of $u$ and $dv$: Always choose $u$ such that its derivative is simpler, and $dv$ such that it is easily integrable. Choosing $u = e^{-x}$ here would make the integral of $dv$ difficult.
• Algebraic errors in integration by parts: Carefully apply the formula and ensure the signs are correct, especially when dealing with the negative sign from integrating $e^{-x}$.
• Forgetting trigonometric identities: The identity $\sec^2 x = 1 + \tan^2 x$ is crucial for relating integrals with different powers of $\tan x$.

Summary This problem is solved by defining a general integral $I(n)$, applying integration by parts to derive a recurrence relation, and then using this relation to simplify the given expression. The key steps involved were the strategic choice of $u$ and $dv$, careful evaluation of limits, and the use of the identity $\sec^2 x = 1 + \tan^2 x$. The recurrence relation $I(n) = -e^{-\pi/4} + n(I(n-1) + I(n+1))$ allowed for the cancellation of terms, leading to a direct numerical answer.

The final answer is $\boxed{50}$ which corresponds to option (B).