Question:
Let
\[
I_1 = \int_{\alpha}^{\pi - \alpha} x \sin x \, dx, I_2 = \int_{\alpha}^{\pi - \alpha} f(\sin x) \, dx
\]
then \( I_2 \) is equal to:
Let
\[
I_1 = \int_{\alpha}^{\pi - \alpha} x \sin x \, dx, I_2 = \int_{\alpha}^{\pi - \alpha} f(\sin x) \, dx
\]
then \( I_2 \) is equal to:
Show Hint
When dealing with integrals involving trigonometric functions, use symmetry and algebraic manipulations to relate the integrals and simplify the calculations.
Updated On: Apr 22, 2026
- \( \frac{\pi}{2} I_1 \)
- \( \pi I_1 \)
- \( \frac{2}{\pi} I_1 \)
- \( 2I_1 \)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Relating the two integrals.
We are given two integrals \( I_1 \) and \( I_2 \). Let us first examine \( I_1 \): \[ I_1 = \int_{\alpha}^{\pi - \alpha} x \sin x \, dx \] We can split the limits of the integral and compute the value of \( I_2 \).
Step 2: Rewriting the integrals.
\( I_2 \) represents the integration of a function of \( \sin x \), so from previous results we have: \[ I_2 = 2 I_1 \]
Step 3: Conclusion.
Thus, \( I_2 \) is equal to \( 2 I_1 \), which matches option (C).
We are given two integrals \( I_1 \) and \( I_2 \). Let us first examine \( I_1 \): \[ I_1 = \int_{\alpha}^{\pi - \alpha} x \sin x \, dx \] We can split the limits of the integral and compute the value of \( I_2 \).
Step 2: Rewriting the integrals.
\( I_2 \) represents the integration of a function of \( \sin x \), so from previous results we have: \[ I_2 = 2 I_1 \]
Step 3: Conclusion.
Thus, \( I_2 \) is equal to \( 2 I_1 \), which matches option (C).
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