Question

The value of \[ \int_0^{\frac{\pi}{2}} \left( \sin x + \sin 2x + \sin 3x \right) dx \] is

• A. \( \frac{8}{3} \)
• B. \( \frac{7}{3} \)
• C. \( \frac{2}{3} \)
• D. 3

Hint:

When integrating trigonometric functions like \( \sin(nx) \), use standard integration formulas and remember to apply the limits correctly.

Answer 1

The Correct Option is B

Step 1: Break down the integral.
We are given: \[ I = \int_0^{\frac{\pi}{2}} \left( \sin x + \sin 2x + \sin 3x \right) dx \] This can be split into three separate integrals: \[ I = \int_0^{\frac{\pi}{2}} \sin x \, dx + \int_0^{\frac{\pi}{2}} \sin 2x \, dx + \int_0^{\frac{\pi}{2}} \sin 3x \, dx \] Step 2: Solve each integral.
Using standard integration formulas, we solve each term: \[ \int_0^{\frac{\pi}{2}} \sin x \, dx = 1, \quad \int_0^{\frac{\pi}{2}} \sin 2x \, dx = 1, \quad \int_0^{\frac{\pi}{2}} \sin 3x \, dx = \frac{1}{3} \] Step 3: Combine the results.
Adding up all the results gives: \[ I = 1 + 1 + \frac{1}{3} = \frac{7}{3} \]