The value of the integral int_{0}^{frac{pi}{2 | vedclass

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(C) Let $I = 60 \int_{0}^{\pi/2} \frac{\sin(6x)}{\sin x} dx$. Using the identity $\sin((2n+1)x) - \sin((2n-1)x) = 2 \cos(2nx) \sin x$,we can express $

Vedclass · Accepted Answer

$104$

Vedclass · Answer

(C) Let $I = 60 \int_{0}^{\pi/2} \frac{\sin(6x)}{\sin x} dx$. Using the identity $\sin((2n+1)x) - \sin((2n-1)x) = 2 \cos(2nx) \sin x$,we can express $\frac{\sin(6x)}{\sin x}$ as a sum of cosines. Specifically,$\frac{\sin(6x)}{\sin x} = \frac{\sin(6x) - \sin(4x) + \sin(4x) - \sin(2x) + \sin(2x)}{\sin x} = 2\cos(5x) + 2\cos(3x) + 2\cos(x)$. Now,integrate each term: $I = 60 \int_{0}^{\pi/2} (2\cos(5x) + 2\cos(3x) + 2\cos(x)) dx$. $I = 60 \left[ \frac{2}{5}\sin(5x) + \frac{2}{3}\sin(3x) + 2\sin(x) ight]_{0}^{\pi/2}$. Evaluating at the limits: $I = 60 \left( (\frac{2}{5}\sin(\frac{5\pi}{2}) + \frac{2}{3}\sin(\frac{3\pi}{2}) + 2\sin(\frac{\pi}{2})) - (0) ight)$. Since $\sin(\frac{5\pi}{2}) = 1$,$\sin(\frac{3\pi}{2}) = -1$,and $\sin(\frac{\pi}{2}) = 1$: $I = 60 \left( \frac{2}{5}(1) + \frac{2}{3}(-1) + 2(1) ight) = 60 \left( \frac{2}{5} - \frac{2}{3} + 2 ight)$. $I = 60 \left( \frac{6 - 10 + 30}{15} ight) = 60 \left( \frac{26}{15} ight) = 4 	imes 26 = 104$.

The value of the integral $\int_{0}^{\frac{\pi}{2}} 60 \frac{\sin (6 x)}{\sin x} d x$ is equal to.

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