Question

Evaluate the definite integral. $$ \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \frac{x}{1+\sin x} \, dx = $$

• A. \(\pi(\sqrt{3} - 2)\)
• B. \(\pi(2 - \sqrt{3})\)
• C. \(\pi(\sqrt{3} + 2)\)
• D. \(\pi/2(2 - \sqrt{3})\)
• E. None of these

Hint:

For any definite integral, the three essential parts are:
• integrand,
• variable,
• limits. e} If even one of these is missing, the answer cannot be found uniquely.

Answer 1

The Correct Option is B

Let $$ I = \int_{\pi/3}^{2\pi/3} \frac{x}{1+\sin x} \, dx \quad \text{... (i)} $$

By using property $$ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx $$

We get,

$$ \begin{aligned} I &= \int_{\pi/3}^{2\pi/3} \frac{\pi - x}{1 + \sin(\pi - x)} \, dx \\ I &= \int_{\pi/3}^{2\pi/3} \frac{\pi - x}{1 + \sin x} \, dx \quad \text{... (ii)} \end{aligned} $$

Adding (i) and (ii), we get

$$ \begin{aligned} 2I &= \int_{\pi/3}^{2\pi/3} \frac{x}{1 + \sin x} + \frac{\pi - x}{1 + \sin x} \, dx \\ &= \int_{\pi/3}^{2\pi/3} \frac{\pi}{1 + \sin x} \, dx \\ &= \pi \int_{\pi/3}^{2\pi/3} \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} \, dx \\ &= \pi \int_{\pi/3}^{2\pi/3} \frac{1 - \sin x}{\cos^2 x} \, dx \\ &= \pi \int_{\pi/3}^{2\pi/3} \left( \sec^2 x - \sec x \tan x \right) dx \\ &= \pi \left[ \tan x - \sec x \right]_{\pi/3}^{2\pi/3} \\ &= \pi \left[ \left( \tan \frac{2\pi}{3} - \sec \frac{2\pi}{3} \right) - \left( \tan \frac{\pi}{3} - \sec \frac{\pi}{3} \right) \right] \\ &= \pi \left[ (-\sqrt{3} + 2) - (\sqrt{3} - 2) \right] \\ &= \pi (-2\sqrt{3} + 4) \\ 2I &= 2\pi(2 - \sqrt{3}) \end{aligned} $$

$$ \therefore \quad I = \pi(2 - \sqrt{3}) $$