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प्रश्न
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"`
उत्तर
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)`
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