Question

The value of the integral $\int_{-\pi}^{\pi} (\sin x + 2 \sin 2x + 3 \sin 3x + 4 \sin 4x + 5 \sin 5x)^2 dx$ is

• A. 101 $\pi$
• B. 100 $\pi$
• C. 55 $\pi$
• D. 50 $\pi$

Answer 1

Answer: (C)

55 \pi

Answer 2

The integral involves the square of a sum of sine functions. Due to the orthogonality property of sine functions over the interval $[-\pi, \pi]$, the integral of product terms $\sin(nx)\sin(mx)$ for $n \neq m$ is zero. The integral of squared terms $\sin^2(nx)$ evaluates to $\pi$. Thus, the integral simplifies to the sum of $n^2 \times \pi$ for $n=1$ to $5$.

$\int_{-\pi}^{\pi} \left( \sum_{n=1}^{5} n \sin nx \right)^2 dx = \sum_{n=1}^{5} n^2 \int_{-\pi}^{\pi} \sin^2 nx dx = \sum_{n=1}^{5} n^2 \pi = \pi (1^2+2^2+3^2+4^2+5^2) = \pi (1+4+9+16+25) = 55\pi$.