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Question
Evaluate the following:
`int_0^pi x sin x cos^2x "d"x`
Solution
Let I = `int_0^pi x sin x cos^2x "d"x` ....(i)
I = `int_0^pi (pi - x) sin(pi - x) cos^2 (pi - x) "d"x`
I = `int_0^pi (pi - x) sin x cos^2x "d"x` .....(ii)
Adding (i) and (ii) we get,
2I = `int_0^pi [x sin x cos^2x + (pi - x)sinx cos^2x]"d"x`
2I = `int_0^pi sinx cos^2x * (x + pi - x) "d"x`
2I = `int__0^pi pi sin x cos^2x "d"x`
= `pi int_0^pi sin x cos^2x "d"x`
Put cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Changing the limits, we have
When x = 0
t = cos 0 = 1
When x = `pi`
= cos `pi` = – 1
2I = `pi int_1^(-1) - "t"^2 "dt"`
= `- pi int_1^(-1) "t"^2 "dt"`
2I = `pi int_(-1)^1 "t"^2 "dt"` ....`[int_"a"^"b" "f"(x)"d"x = - int_"b"^"a" "f"(x) "d"x]`
2I = `pi["t"^3/3]_(-1)^1`
= `pi[1/3 + 1/3]`
= `pi(2/3)`
∴ I = `pi/3`
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