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Grade 12Integral Calculus

Integral [sin(nx)]^2/[sin(x)]^2, upper limit : pi, lower limit : 0

Abhishek Ghatge

12 Years agoGrade 12

1 Answer

Jitender Singh

12 Years ago

Ans:

$I_{n} = \int_{0}^{\pi}\frac{sin^{2}(nx)}{sin^{2}(x)}dx$

$I_{1} = \int_{0}^{\pi}\frac{sin^{2}(x)}{sin^{2}(x)}dx$

$I_{1} = \pi$

$I_{2} = \int_{0}^{\pi}\frac{sin^{2}(2x)}{sin^{2}(x)}dx$

$I_{2} = \int_{0}^{\pi}\frac{(2sin(x).cos(x))^{2}}{sin^{2}(x)}dx$

$I_{2} = \int_{0}^{\pi}4cos^{2}(x)dx$ $I_{2} = \int_{0}^{\pi}2(1+cos(2x))dx = (2x + sin(2x))_{0}^{\pi} = 2\pi$

$I_{3} = \int_{0}^{\pi}\frac{sin^{2}(3x)}{sin^{2}(x)}dx$

$I_{3} = (3x + 2sin(2x) + \frac{1}{2}sin (4x))_{0}^{\pi}$

$I_{3} = 3\pi$

$I_{4} = \int_{0}^{\pi}\frac{sin^{2}(4x)}{sin^{2}(x)}dx$

$I_{4} = (4x+3sin(2x) + sin(4x) + \frac{1}{3}sin(6x))_{0}^{\pi}$

$I_{4} =4\pi$

Similarly, when we do for the n, we get

$I_{n} =n\pi$

Thanks & Regards

Jitender Singh

IIT Delhi

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