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Grade: 12
        Integral [sin(nx)]^2/[sin(x)]^2, upper limit : pi, lower limit : 0
6 years ago

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
							
Ans:
I_{n} = \int_{0}^{\pi}\frac{sin^{2}(nx)}{sin^{2}(x)}dx
I_{1} = \int_{0}^{\pi}\frac{sin^{2}(x)}{sin^{2}(x)}dx
I_{1} = \pi
I_{2} = \int_{0}^{\pi}\frac{sin^{2}(2x)}{sin^{2}(x)}dx
I_{2} = \int_{0}^{\pi}\frac{(2sin(x).cos(x))^{2}}{sin^{2}(x)}dx
I_{2} = \int_{0}^{\pi}4cos^{2}(x)dxI_{2} = \int_{0}^{\pi}2(1+cos(2x))dx = (2x + sin(2x))_{0}^{\pi} = 2\pi
I_{3} = \int_{0}^{\pi}\frac{sin^{2}(3x)}{sin^{2}(x)}dx
I_{3} = (3x + 2sin(2x) + \frac{1}{2}sin (4x))_{0}^{\pi}
I_{3} = 3\pi
I_{4} = \int_{0}^{\pi}\frac{sin^{2}(4x)}{sin^{2}(x)}dx
I_{4} = (4x+3sin(2x) + sin(4x) + \frac{1}{3}sin(6x))_{0}^{\pi}
I_{4} =4\pi
Similarly, when we do for the n, we get
I_{n} =n\pi
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
5 years ago
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