Question

The value of $\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$ is equal to

• A. $\frac{7}{2}-\sqrt{3}-\log _e \sqrt{3}$
• B. $\frac{10}{3}-\sqrt{3}-\log _e \sqrt{3}$
• C. $\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3}$
• D. $-2+3 \sqrt{3}+\log _e \sqrt{3}$

Hint:

When solving integrals with trigonometric functions, use standard trigonometric identities like \( \cos x = 2 \cos^2(x/2) - 1 \) and substitutions for easier integration.

Answer 1

The Correct Option is B

We start with the integral:

$\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$

Step 1: Simplify the integrand.

Rewrite the expression:

= 2 ∫ dx / (sinx + sinxcosx) + 3 ∫ dx / (1 + cosx)

Step 2: Solve each integral separately.

The first integral:

∫ dx / (sinx + sinxcosx) = ∫ (cosecx - cotx) cosecxc dx from π/3 to π/2

The second integral:

= ∫ (cosec x - cot x) dx = (1) from π/3 to π/2

Step 3: Combine the results.

2 * (1 - 1/√3) = 1 - 1/√3

Step 4: Solve the remaining part of the integral:

= ∫ dx / (sinx + cosx)

Use substitution and evaluate using standard methods for trigonometric integrals.

Final result:

10/3 - √3 + ln(√3)