Question

Evaluate the following:

`int_0^pi x sin x cos^2x "d"x`

Answer 1

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`