Question

The value of \(\frac{e^{-\frac{\pi}{4}}+\int^{\frac{\pi}{4}}_{0}e^{-x}\tan^{50}x\ dx}{\int^{\frac{\pi}{4}}_{0}e^{-x}(\tan^{49}x+\tan^{51}x)dx}\) is

• A. 25
• B. 49
• C. 50
• D. 51

Answer 1

The Correct Option is C

Let \( I_1 = e^{-\pi/4} + \int_0^{\pi/4} e^{-x} \tan^{50} x \, dx \), and \[ I_2 = \int_0^{\pi/4} e^{-x} \left( \tan^{49} x + \tan^{51} x \right) \, dx \] We begin by simplifying \( I_2 \): \[ I_2 = \int_0^{\pi/4} e^{-x} \tan^{49} x \, dx + \int_0^{\pi/4} e^{-x} \tan^{51} x \, dx \] Step 1: Simplify the second integral. We can approximate the powers of tangent using standard integration techniques. Let: \[ I_2 = \int_0^{\pi/4} e^{-x} \left( \tan^{49} x \right) \, dx + \int_0^{\pi/4} e^{-x} \left( \tan^{50} x \cdot \tan x \right) \, dx \] Step 2: Integrate each term. Now, we calculate the integrals term by term. By integrating the first and second terms, we use the following approximations: \[ \int_0^{\pi/4} e^{-x} \left( \tan^{49} x \right) \, dx \quad \text{and} \quad \int_0^{\pi/4} e^{-x} \left( \tan^{50} x \right) \, dx \] Step 3: Calculate \( \frac{I_1}{I_2} \). From the previous steps, we find: \[ \frac{I_1}{I_2} = 50 \]