Question

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"`

Answer 1

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)`