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प्रश्न
Evaluate: `int_0^pi ("x"sin "x")/(1+ 3cos^2 "x") d"x"`.
उत्तर
Let `"I" = int_0^pi ("x"sin "x")/(1+ 3cos^2 "x") d"x"` ...(i)
⇒ `"I" = int_0^pi ((pi-"x")sin(pi-"x"))/(1+3cos^2(pi-"x"))d"x"`
= `int_0^pi (pisin"x")/(1+3cos^2"x")d"x" - int_0^pi (xsin"x")/(1+3cos^2"x")d"x"` ...(ii)
Adding (i) & (ii), we have
we get: `2"I" = int_0^pi(pisin"x")/(1+3 cos^2 "x")` dx
Put cos x = t
⇒ - sin x dx = dt, when x = 0
⇒ t = 1, for x = π ⇒ t = - 1
So, `2I = π int_1^-1 dt/(1 + 3t^2)`
⇒ `π/3 int_-1^1 (dt)/((1/sqrt3)^2 + (t)^2)`
⇒ `π/3 xx sqrt3 [tan^-1(sqrt3t)]_-1^1`
⇒ `(sqrt3π)/3 [ tan^-1sqrt3 - ( - tan^-1 sqrt3)]`
I = `(sqrt3π)/3. π/3 = sqrt3π^2/9`
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