Question

The value of the integral \(\int_{0}^{3\pi/2} \frac{\cos^3 x}{\cos^3 x + \sin^3 x} \, dx\) is:

• A. 0
• B. 1
• C. $\frac{\pi}{4}$
• D. $\frac{3\pi}{4}$

Hint:

For any integral of the form $\int_{0}^{a} \frac{f(x)}{f(x) + f(a-x)} \, dx$, the result is simply half of the upper limit: $\frac{a}{2}$. In this problem, $\frac{3\pi/2}{2} = \frac{3\pi}{4}$.

Answer 1

The Correct Option is D

Concept: This integral can be solved efficiently using the "King's Property" of definite integrals: $$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$$

Step 1: Apply the property.
Let $I = \int_{0}^{3\pi/2} \frac{\cos^3 x}{\cos^3 x + \sin^3 x} \, dx$ Replacing $x$ with $(\frac{3\pi}{2} - x)$: $$I = \int_{0}^{3\pi/2} \frac{\cos^3 (\frac{3\pi}{2} - x)}{\cos^3 (\frac{3\pi}{2} - x) + \sin^3 (\frac{3\pi}{2} - x)} \, dx$$ Using the trigonometric identities $\cos(\frac{3\pi}{2} - x) = -\sin x$ and $\sin(\frac{3\pi}{2} - x) = -\cos x$: $$I = \int_{0}^{3\pi/2} \frac{(-\sin x)^3}{(-\sin x)^3 + (-\cos x)^3} \, dx$$ $$I = \int_{0}^{3\pi/2} \frac{-\sin^3 x}{-(\sin^3 x + \cos^3 x)} \, dx = \int_{0}^{3\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} \, dx$$

Step 2: Combine the integrals.
Adding the two expressions for $I$: $$2I = \int_{0}^{3\pi/2} \left( \frac{\cos^3 x + \sin^3 x}{\cos^3 x + \sin^3 x} \right) dx$$ $$2I = \int_{0}^{3\pi/2} 1 \, dx$$

Step 3: Final Calculation.
$$2I = [x]_{0}^{3\pi/2}$$ $$2I = \frac{3\pi}{2} - 0 \implies I = \frac{3\pi}{4}$$ The correct choice is (d).