Question

Evaluate the integral: $\int_0^{\pi / 2} \frac{\cos x}{3 \cos x + \sin x} dx$

• A. $\frac{3\pi}{20} - \frac{1}{10}\log 3$
• B. $\frac{3\pi}{10} - \log 3$
• C. $\frac{3\pi}{20} + \log 3$
• D. $\frac{3\pi}{20} - \log 3$

Hint:

For any definite integral of this form from $0$ to $\frac{\pi}{2}$, the first coefficient $A$ directly determines the linear multiplier of the interval width. Because $A = \frac{3}{10}$ and the width is $\frac{\pi}{2}$, the first term of your answer must be $\frac{3}{10} \times \frac{\pi}{2} = \frac{3\pi}{20}$, allowing you to easily rule out option (B).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires us to calculate the value of a definite integral involving trigonometric expressions in both the numerator and the denominator from $0$ to $\frac{\pi}{2}$.

Step 2: Key Formula or Approach:
For integrals of the form $\int \frac{p \cos x + q \sin x}{a \cos x + b \sin x} dx$, express the numerator as a linear combination of the denominator and its derivative: $$\text{Numerator} = A(\text{Denominator}) + B\left(\frac{d}{dx}(\text{Denominator})\right)$$ Once the constants $A$ and $B$ are determined, the integration splits into a simple constant integral and a logarithmic function integral ($\int \frac{f'(x)}{f(x)} dx = \log|f(x)|$).

Step 3: Detailed Explanation:
Let the numerator be expressed as: $$\cos x = A(3 \cos x + \sin x) + B\left(\frac{d}{dx}(3 \cos x + \sin x)\right)$$ $$\cos x = A(3 \cos x + \sin x) + B(-3 \sin x + \cos x)$$ Rearranging terms by grouping $\cos x$ and $\sin x$: $$\cos x = (3A + B)\cos x + (A - 3B)\sin x$$ Equating the coefficients of $\cos x$ and $\sin x$ on both sides: $$3A + B = 1 \quad \text{--- (1)}$$ $$A - 3B = 0 \implies A = 3B \quad \text{--- (2)}$$ Substitute (2) into (1): $$3(3B) + B = 1 \implies 10B = 1 \implies B = \frac{1}{10}$$ Using $A = 3B$: $$A = \frac{3}{10}$$ Now substitute these values back into our integral $I$: $$I = \int_0^{\pi / 2} \frac{\frac{3}{10}(3 \cos x + \sin x) + \frac{1}{10}(-3 \sin x + \cos x)}{3 \cos x + \sin x} dx$$ $$I = \frac{3}{10} \int_0^{\pi / 2} 1 \cdot dx + \frac{1}{10} \int_0^{\pi / 2} \frac{-3 \sin x + \cos x}{3 \cos x + \sin x} dx$$ $$I = \frac{3}{10} \Big[ x \Big]_0^{\pi/2} + \frac{1}{10} \Big[ \log|3 \cos x + \sin x| \Big]_0^{\pi/2}$$ Evaluating the limits: $$I = \frac{3}{10}\left(\frac{\pi}{2} - 0\right) + \frac{1}{10}\Big( \log|3\cos(\pi/2) + \sin(\pi/2)| - \log|3\cos(0) + \sin(0)| \Big)$$ Since $\cos(\pi/2) = 0$, $\sin(\pi/2) = 1$, $\cos(0) = 1$, and $\sin(0) = 0$: $$I = \frac{3\pi}{20} + \frac{1}{10}\Big( \log(1) - \log(3) \Big)$$ Since $\log(1) = 0$: $$I = \frac{3\pi}{20} - \frac{1}{10}\log 3$$

Step 4: Final Answer:
The value of the integral is $\frac{3\pi}{20} - \frac{1}{10}\log 3$, which corresponds to option (A).