# Prove that ∫ a 0 F ( X ) D X = ∫ a 0 F ( a − X ) D X , Hence Evaluate ∫ π 0 X Sin X 1 + Cos 2 X D X - Mathematics

Sum

Prove that int_0^"a" "f" ("x") "dx" = int_0^"a" "f" ("a" - "x") "d x", hence evaluate int_0^pi ("x" sin "x")/(1 + cos^2 "x") "dx"

#### Solution

To prove: int_0^a "f"("x") "dx" = int_0^a "f" ("a - x") "dx"

Proof: Let t = a - x
⇒ dt = - dx
When x = 0, t = a
When x = a , t = 0
Putting the value of x in LHS

int_a^0 "f"("a - t") (- "dt")

= - int_a^0 "f" ("a - t") ("dt")

= int_0^a "f" ("a - t") ("dt")

= int_0^a ("a - x") ("dx")      ...(∵ int_a^b "f" (t) "dt" = int_a^b ("x")( "dx"))
= RHS

Using this we can solve the given question as follows:

I = int_0^pi f ("x") d"x" = int_0^pi (pi - "x") "dx"

⇒ 2I = int_0^pi f ("x") d"x" + int_0^pi f (pi - "x") d"x" = int_0^pi ("x" sin "x")/(1 + cos^2 "x") d"x" + int_0^pi ((pi - "x") sin(pi - "x"))/(1 + cos^2 (pi - "x")) d"x"

⇒2"I" = int_0^pi ("x" sin "x")/(1 + cos^2 "x") "dx" + int_0^pi ((pi - "x")sin"x")/(1 + cos^2 (pi - "x")) "dx"

⇒ 2"I" = int_0^pi (pi sin"x")/(1 + cos^2 "x") "dx"

Let, cos x = t ⇒ -sin x dx = dt

⇒ 2"I" = -int_1^-1 (pi)/(1 + t^2) dt = -pi [ tan^-1 t ]_1^(-1) = -pi(-pi/(4) - pi/(4)) = pi^2/(2)

∴ "I" = pi^2/(4)

Concept: Properties of Definite Integrals
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