Question

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.

• A. $\pi$
• B. $\frac{3\pi}{2}$
• C. $\frac{\pi}{2}$
• D. $\frac{\pi}{4}$
• E. $2\pi$

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!

Answer 1

The Correct Option is D

Concept:
This problem uses the King's Rule of definite integration, which states: $$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$$ By adding the original integral to the transformed integral, the integrand often simplifies significantly, particularly with complementary trigonometric functions.
Step 1: Label the original integral.
Let $I$ be the given integral: $$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{11}x}{\cos^{11}x + \sin^{11}x} dx \quad \text{--- (Equation 1)}$$
Step 2: Apply the definite integral property.
Substitute $x$ with $(\frac{\pi}{2} - x)$: $$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{11}(\frac{\pi}{2}-x)}{\cos^{11}(\frac{\pi}{2}-x) + \sin^{11}(\frac{\pi}{2}-x)} dx$$ Since $\cos(\frac{\pi}{2}-x) = \sin x$ and $\sin(\frac{\pi}{2}-x) = \cos x$, the integral becomes: $$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{11}x}{\sin^{11}x + \cos^{11}x} dx \quad \text{--- (Equation 2)}$$
Step 3: Add Equation 1 and Equation 2.
$$I + I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{11}x}{\cos^{11}x + \sin^{11}x} dx + \int_{0}^{\frac{\pi}{2}} \frac{\sin^{11}x}{\sin^{11}x + \cos^{11}x} dx$$ Combine the numerators over the common denominator: $$2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{11}x + \sin^{11}x}{\cos^{11}x + \sin^{11}x} dx$$
Step 4: Evaluate the simplified integral.
The integrand cancels out perfectly to $1$: $$2I = \int_{0}^{\frac{\pi}{2}} 1 dx$$ $$2I = [x]_{0}^{\frac{\pi}{2}}$$ $$2I = \frac{\pi}{2} - 0$$ Divide by 2 to solve for $I$: $$I = \frac{\pi}{4}$$