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: Set up the splitting trick.
Write the numerator as $A(\text{denominator}) + B(\text{derivative of denominator})$: $\cos x = A(3\cos x + \sin x) + B(-3\sin x + \cos x)$.
Step 2: Match coefficients.
Comparing $\cos x$ terms: $3A + B = 1$. Comparing $\sin x$ terms: $A - 3B = 0$, so $A = 3B$.
Step 3: Solve for A and B.
Substituting, $3(3B) + B = 1 \Rightarrow 10B = 1 \Rightarrow B = \tfrac{1}{10}$ and $A = \tfrac{3}{10}$.
Step 4: Break the integral in two.
$I = \dfrac{3}{10}\displaystyle\int_0^{\pi/2} dx + \dfrac{1}{10}\displaystyle\int_0^{\pi/2} \dfrac{-3\sin x + \cos x}{3\cos x + \sin x}\,dx$.
Step 5: Integrate each part.
The first gives $\dfrac{3}{10}\cdot\dfrac{\pi}{2} = \dfrac{3\pi}{20}$. The second is of the form $\int \tfrac{f'}{f}$, giving $\dfrac{1}{10}\big[\log|3\cos x + \sin x|\big]_0^{\pi/2}$.
Step 6: Plug in the limits.
At $\tfrac{\pi}{2}$ the bracket is $\log 1 = 0$; at $0$ it is $\log 3$. So the second part is $\dfrac{1}{10}(0 - \log 3) = -\dfrac{1}{10}\log 3$. Hence $I = \dfrac{3\pi}{20} - \dfrac{1}{10}\log 3$, option 1, matching the key.
\[ \boxed{\dfrac{3\pi}{20} - \tfrac{1}{10}\log 3} \]