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Question
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Solution
let a + b - x = t
⇒ dx = -dt
when x = a,t = b and x = b,t = a
`int_a^b ƒ("x") d"x" = -int_b^aƒ(a + b -"t")d"t"`
= `int_a^bƒ(a + b -"t")d"t" ...[∵ int_a^b ƒ("x") d"x" = -int_b^a ƒ("x") d"x"]`
= `int_a^bƒ(a + b -"x")d"x" ...[∵ int_a^b ƒ("x") d"x" = int_a^b ƒ("t") d"t"]`
Hence proved.
let `I = int_(π/6)^(π/3) (d"x")/(1+ sqrt(tan "x")) = int_(π/6)^(π/3)(sqrt(cos"x")d"x")/(sqrt(cos"x")+ sqrt(sin"x"))` .....(ii)
Then, using the property from (i)
`I = int_(π/6)^(π/3) (sqrtcos(π/3 + π/6 - "x") d"x")/ (sqrtcos(π/3 + π/6 - "x") + sqrtsin(π/3 + π/6 - "x"))`
= `int_(π/6)^(π/3) (sqrt(sin"x")d"x")/(sqrt(sin"x") + sqrt(cos"x")` ......(iii)
Adding (ii) and (iii), we get
`2I = int_(π/6)^(π/3)d"x" = ["x"](π/3)/(π/6) = π/3 - π/6 = π/6`
⇒ `I = π/12`
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