Question

The integral \(\int_{0}^{\pi/4} \frac{136 \sin x}{3 \sin x + 5 \cos x} \, dx\) is equal to

• A. \( 3\pi - 50 \log_e 2 + 20 \log_e 5 \)
• B. \( 3\pi - 25 \log_e 2 + 10 \log_e 5 \)
• C. \( 3\pi - 10 \log_e (2\sqrt{2}) + 10 \log_e 5 \)
• D. \( 3\pi - 30 \log_e 2 + 20 \log_e 5 \)

Answer 1

The Correct Option is A

Let \[ I = \int_{0}^{\pi / 4} \frac{136 \sin x}{3 \sin x + 5 \cos x} \, dx \] We can express \(136 \sin x\) in terms of \(3 \sin x + 5 \cos x\) and \(3 \cos x - 5 \sin x\):

\[ 136 \sin x = A (3 \sin x + 5 \cos x) + B (3 \cos x - 5 \sin x) \] This gives the equations:

\[ 136 = 3A - 5B \quad \dots (1) \] \[ 0 = 5A + 3B \quad \dots (2) \]

From equation (2), we have \(B = -\frac{5}{3} A\). Substitute \(B = -\frac{5}{3} A\) into equation (1):

\[ 136 = 3A - 5 \left(-\frac{5}{3} A \right) \] \[ 136 = 3A + \frac{25}{3} A \] \[ 136 = \frac{34A}{3} \] \[ A = \frac{136 \times 3}{34} = 12 \] \[ B = -\frac{5}{3} (12) = -20 \]

Now, rewrite \(I\) as:

\[ I = \int_{0}^{\pi/4} A \frac{(3 \sin x + 5 \cos x)}{3 \sin x + 5 \cos x} \, dx + \int_{0}^{\pi/4} B \frac{(3 \cos x - 5 \sin x)}{3 \sin x + 5 \cos x} \, dx \] \[ = \int_{0}^{\pi/4} A \, dx + B \left[ \ln (3 \sin x + 5 \cos x) \right]_{0}^{\pi/4} \] Substitute \(A = 12\) and \(B = -20\):

\[ = 12 \left(\frac{\pi}{4}\right) - 20 \left[\ln \left(\frac{3}{\sqrt{2}} + \frac{5}{\sqrt{2}} \right) - \ln (0 + 5)\right] \] \[ = 3\pi - 20 \ln \left(\frac{8}{\sqrt{2}}\right) + 20 \ln 5 \] \[ = 3\pi - 20 \ln (4\sqrt{2}) + 20 \ln 5 \] \[ = 3\pi - 20 \left(\ln 4 + \ln \sqrt{2}\right) + 20 \ln 5 \] \[ = 3\pi - 20 (\ln 2^2 + \ln 2^{1/2}) + 20 \ln 5 \] \[ = 3\pi - 50 \ln 2 + 20 \ln 5 \]

Answer 2

Step 1: Use the substitution t = tan x
Let t = tan x. Then dx = dt/(1 + t^2), sin x = t/√(1 + t^2), cos x = 1/√(1 + t^2). On [0, π/4], the limits become t ∈ [0, 1].
The denominator transforms as \[ 3\sin x + 5\cos x = \frac{3t + 5}{\sqrt{1 + t^2}}. \] The numerator times dx becomes \[ 136\sin x\,dx = 136\cdot \frac{t}{\sqrt{1 + t^2}} \cdot \frac{dt}{1 + t^2}. \] Therefore, \[ \int_{0}^{\pi/4}\frac{136\sin x}{3\sin x + 5\cos x}\,dx = \int_{0}^{1} \frac{136\,t}{(3t + 5)(1 + t^2)}\,dt. \]

Step 2: Partial-fraction decomposition
Decompose \[ \frac{t}{(3t + 5)(1 + t^2)} = \frac{A}{3t + 5} + \frac{Bt + C}{1 + t^2}. \] Equating coefficients gives \[ A = -\frac{15}{34},\qquad B = \frac{5}{34},\qquad C = \frac{3}{34}. \] Thus \[ \frac{t}{(3t + 5)(1 + t^2)} = -\frac{15}{34}\cdot\frac{1}{3t + 5} + \frac{5}{34}\cdot\frac{t}{1 + t^2} + \frac{3}{34}\cdot\frac{1}{1 + t^2}. \]

Step 3: Integrate term by term and restore the factor 136
\[ \int \frac{1}{3t + 5}\,dt = \frac{1}{3}\ln(3t + 5),\quad \int \frac{t}{1 + t^2}\,dt = \frac{1}{2}\ln(1 + t^2),\quad \int \frac{1}{1 + t^2}\,dt = \arctan t. \] Multiplying by 136, we obtain the antiderivative \[ -20\ln(3t + 5) + 10\ln(1 + t^2) + 12\arctan t. \] Therefore, \[ \int_{0}^{1} \frac{136\,t}{(3t + 5)(1 + t^2)}\,dt = \Big[-20\ln(3t + 5) + 10\ln(1 + t^2) + 12\arctan t\Big]_{0}^{1}. \]

Step 4: Evaluate at the bounds and simplify
At t = 1: \[ -20\ln(8) + 10\ln(2) + 12\cdot\frac{\pi}{4} = -20\ln(8) + 10\ln(2) + 3\pi. \] At t = 0: \[ -20\ln(5) + 10\ln(1) + 12\cdot 0 = -20\ln(5). \] Subtracting, \[ \int_{0}^{\pi/4}\frac{136\sin x}{3\sin x + 5\cos x}\,dx = \big(-20\ln 8 + 10\ln 2 + 3\pi\big) - \big(-20\ln 5\big) = 3\pi - 20\ln 8 + 10\ln 2 + 20\ln 5. \] Use \(\ln 8 = 3\ln 2\) to get \[ 3\pi - 20\cdot 3\ln 2 + 10\ln 2 + 20\ln 5 = 3\pi - 50\ln 2 + 20\ln 5. \]

Final answer

\( 3\pi - 50 \log_e 2 + 20 \log_e 5 \)