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प्रश्न
By using the properties of the definite integral, evaluate the integral:
`int_0^pi (x dx)/(1+ sin x)`
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उत्तर
Let `I = int_0^pi (x dx)/ (1 + sin x)`
`I = int_0^pi (pi - x)/ (1 + sin (pi-x)) dx` `...[∵ int_0^a f (x) dx = int_0^a f (a - x) dx]`
`= int_0^pi (pi - x)/ (1 + sin x) dx`
Adding (i) and (ii), we get
`2 I = int_0^pi (x + pi - x)/ (1 + sin x) dx`
`= pi int_0^pi 1/ (1 + sin x) dx`
`= pi int_0^pi (1 - sin x)/ (1 - sin^2 x) dx`
`= pi int_0^pi (1 - sin x)/(cos^2 x) dx`
`= pi int_0^pi (sec^2 x - tan x sec x) dx`
`= pi [tan x - sec x]_0^pi`
= π [(tan π - sec π) - (tan0 - sec0)]
= π [(0 - (-1)) - (0 - 1)]
= 2π
Hence, I = π
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