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
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
Updated On: Feb 27, 2026
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The Correct Option is C
Solution and Explanation
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:
\[
\frac{I_1}{I_2} = 50
\]
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