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 \(B = -20\):

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