Question:medium
The value of $\int_{0}^{\frac{\pi}{2}}\frac{\cos^{11}x}{\cos^{11}x+\sin^{11}x}dx$ is equal to}
\textit{Note: The original exam paper contained a typographical error ($c\hat{D}$ instead of $dx$). It has been corrected here to permit a valid solution.
The value of $\int_{0}^{\frac{\pi}{2}}\frac{\cos^{11}x}{\cos^{11}x+\sin^{11}x}dx$ is equal to}
\textit{Note: The original exam paper contained a typographical error ($c\hat{D}$ instead of $dx$). It has been corrected here to permit a valid solution.
Show Hint
Logic Tip: Any definite integral of the form $\int_{0}^{\pi/2} \frac{f(\sin x)}{f(\sin x) + f(\cos x)} dx$ will always evaluate to exactly half of the upper limit, which is $\frac{\pi}{4}$. The power $11$ is purely a distractor!
Updated On: Apr 27, 2026
- $\pi$
- $\frac{3\pi}{2}$
- $\frac{\pi}{2}$
- $\frac{\pi}{4}$
- $2\pi$
Show Solution
The Correct Option is D
Solution and Explanation
Was this answer helpful?
0
Top Questions on Some Properties of Definite Integrals
- If the solution of \[ \left( 1 + 2e^\frac{x}{y} \right) dx + 2e^\frac{x}{y} \left( 1 - \frac{x}{y} \right) dy = 0 \] is \[ x + \lambda y e^\frac{x}{y} = c \quad \text{(where \(c\) is an arbitrary constant), then \( \lambda \) is:} \]
- VITEEE - 2024
- Mathematics
- Some Properties of Definite Integrals
- Let \( f(x) = \int_0^x g(t) \log_e \left( \frac{1 - t}{1 + t} \right) dt \), where \( g \) is a continuous odd function. If \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( f(x) + \frac{x^2 \cos x}{1 + e^x} \right) dx = \left( \frac{\pi}{\alpha} \right)^2 - \alpha, \] then \( \alpha \) is equal to .....
- JEE Main - 2024
- Mathematics
- Some Properties of Definite Integrals
- The value of \( \displaystyle \int_0^{1.5} \left\lfloor x^2 \right\rfloor \, dx \) is:
- WBJEE - 2025
- Mathematics
- Some Properties of Definite Integrals
- The value of \( \int_{-100}^{100} \frac{x + x^3 + x^5}{1 + x^2 + x^4 + x^6} dx \) is:
- WBJEE - 2025
- Mathematics
- Some Properties of Definite Integrals
- Want to practice more? Try solving extra ecology questions todayView All Questions
