Question

y suffix n = intregation sinnx/sinx dx
then show that
y= 2sin (n-1)x/(n-1) +y suffix n-2
HENCE evaluate
intregation limit 0 to pie /2
sin7x/sinx

Answer 1

Answer on Question #45493 – Math - Integral Calculus

y suffix n = intregation sinnx/sinx dx

then show that

$$y = 2 \sin (n-1)x / (n-1) + y \text{ suffix } n-2$$

HENCE evaluate

intregation limit 0 to pie /2

sin7x/sinx

Solution.

$$\begin{aligned} y &= \text{suffix } (n) = \int \frac{\sin nx}{\sin x} dx = \int \frac{\sin[(n-2)x + 2x]}{\sin x} dx = \\ &= \int \frac{\sin(n-2)x * (1 - 2\sin^2 x) + 2\cos(n-2)x * \sin x * \cos x}{\sin x} dx = \\ &= \text{suffix } (n-2) + \int -2\sin(n-2)x * \sin x + 2\cos(n-2)x * \cos x dx = \\ &= \text{suffix } (n-2) - 2\int \cos(n-1)x dx = \frac{2\sin(n-1)x}{n-1} + \text{suffix } (n-2). \end{aligned}$$

So,

$$\begin{aligned} \text{suffix}(7) = \text{suffix}(5) + \frac{2\sin6x}{6} = \text{suffix}(3) + \frac{2\sin4x}{4} + \frac{\sin6x}{3} = \\ = \text{suffix}(1) + \frac{2\sin2x}{2} + \frac{2\sin4x}{4} + \frac{\sin6x}{3} = \int dx + \sin2x + \frac{\sin4x}{2} + \frac{\sin6x}{3} = \\ = x + \sin2x + \frac{\sin4x}{2} + \frac{\sin6x}{3} + \text{const}. \end{aligned}$$

Thus: $\int_0^{\frac{\pi}{2}}\frac{\sin7x}{\sin x} dx = \left(x + \sin2x + \frac{\sin4x}{2} + \frac{\sin6x}{3}\right)\big|_x^{x = \frac{\pi}{2}} = \frac{\pi}{2}.$

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