# Prove that ∫ B a ƒ ( X ) D X = ∫ B a ƒ ( a + B − X ) D X and Hence Evaluate ∫ π 3 π 6 D X 1 + √ Tan X - Mathematics

#### Question

Sum

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

Concept: Definite Integrals Problems
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