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 $B = -20$:

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