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Question
Evaluate the following integrals using properties of integration:
`int_0^(2pi) x log((3 + cosx)/(3 - cosx)) "d"x`
Solution
Let I = `int_0^(2pi) x log((3 + cosx)/(3 - cosx)) "d"x`
Let f(x) = `log((3 + cos x)/(3 - cos x))`
`"f"(2pi - x) = log((3 + cos(2pi - x))/(3 - cos(2pi - x)))`
= `log((3 + cosx)/(3 - cosx))`
= f(x)
∵ `int_0^"a" x f(x) "d"x = "a"/2 int_0^"a" f(x) "d"x "if" "f"("a" - x)` = f(x)
I = `(2pi)/2 int_0^(2pi) log((3 + cos x)/(3 - cos x)) "d"x`
= `2pi int_0^pi log ((3 + cos x)/(3 - cos x)) "d"x` .......(1)
I = `2pi int_0^pi log ((3 + cos (pi - x))/(3 - cos(pi - x)) "d"x`
`int_0^"a" f(x) "d"x = int_0^"a" f("a" - x) "d"x`
= `2 pi int_0^pi log((3 - cosx)/(3 + cos x)) "d"x` ........(2)
Add (1) and (2)
2I = `2pi int_0^pi (log((3 + cosx)/(3 - cosx)) + log((3 - cosx)/(3 + cosx)))"d"x`
= `2pi int_0^pi log((3 + cosx)/(3 - cosx) * (3 - cosx)/(3 + cosx)) "d"x`
2I = `2pi xx 0` 0
I = 0
`int_0^(2pi) x log ((3 cos x)/(3 - cos x)) "d"x` = 0
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