Question

\[ \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} d x= \]

• A. \(\pi (\sqrt{3}-2)\)
• B. \(\pi (2-\sqrt{3})\)
• C. \(\pi (\sqrt{3}+2)\)
• D. \(\frac{\pi}{2}(2-\sqrt{3})\)

Hint:

Use the property that replaces x by (a + b - x), then integrate 1/(1 + sin x).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The limits are \(\dfrac\pi3\) and \(\dfrac{2\pi}3\), with sum \(\pi\). Use \(\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\).

Step 2: Apply the property
\(I=\int\dfrac{x}{1+\sin x}dx=\int\dfrac{\pi-x}{1+\sin x}dx\) since \(\sin(\pi-x)=\sin x\). Adding: \(2I=\pi\int_{\pi/3}^{2\pi/3}\dfrac{dx}{1+\sin x}\).

Step 3: Integrate
\(\int\dfrac{dx}{1+\sin x}=\int\dfrac{1-\sin x}{\cos^{2}x}dx=\tan x-\sec x\).

Step 4: Apply the limits
At \(\dfrac{2\pi}3\) the value is \(-\sqrt3+2\). At \(\dfrac\pi3\) it is \(\sqrt3-2\). The difference is \(2(2-\sqrt3)\), so \(2I=2\pi(2-\sqrt3)\).

Step 5: Result
\(I=\pi(2-\sqrt3)\).

Final Answer:
The value of the integral is pi(2 - sqrt3). \[ \boxed{\text{(B) }\pi(2-\sqrt{3})} \]