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I need help on this question....pls...help.....how to perform this integration...

I need help on this question....pls...help.....how to perform this integration...

Question Image
Grade:11

1 Answers

RAHUL ROBIN
16 Points
3 years ago
Let In = 1/pi int((0)to(pi/2)) sin^2(nx)/sin^2(x) dx
In - In-1 = 1/pi int((0)to(pi/2)) (sin^2(nx)-sin^2((n-1)x))/sin^2(x) dx
In - In-1 = 1/pi int((0)to(pi/2)) (cos2(n-1)x-cos2nx)/2sin^2(x) dx
In - In-1 = 1/pi int((0)to(pi/2)) 2sinxsin(2n-1)x/2sin^2(x) dx
In - In-1 = 1/pi int((0)to(pi/2)) sin(2n-1)x/sinx dx
In-1 - In-2 = 1/pi int((0)to(pi/2)) sin(2n-3)x/sinx dx
In - 2In-1 + In-2 = 1/pi int((0)to(pi/2)) (sin(2n-1)x-sin(2n-3)x)/sinx dx
In - 2In-1 + In-2 = 1/pi int((0)to(pi/2)) 2sinxcos(2n-2)x/sinx dx
In -2In-1 + In-2 = 1/pi int((0)to(pi/2)) 2cos(2n-2)x dx
In - 2In-1 + In-2 = 1/pi (sin(2n-2)x/(n-1)) (0)to(pi/2) = 0
In + In-2 = 2In-1
forms an A.P
I1 = 1/pi int((0)to(pi/2)) sin^2(x)/sin^2(x) dx =0.5
I2 = 1/pi int((0)to(pi/2)) sin^2(2x)/sin^2(x) dx 
     = 1/pi int((0)to(pi/2)) 4sin^2(x)cos^2(x)/sin^2(x) dx = 1
d= I2-I1 = 1-0.5 =0.5
In = I1 + d(n-1) = n/2
 

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