Question

Evaluate the following : `int_0^pi x*sinx*cos^4x*dx`

Answer 1

Let I = `int_0^pi x*sinx*cos^4x*dx`              ...(1)

We use the property, `int_0^a f(x)*dx = int_0^a f(a - x)*dx`

Here a = π.
Hence changing x by π – x, we get

I = `int_0^pi (pi - x)*sin(pi - x)*[cos(pi - x)]^4*dx`

= `int_0^pi (pi - x)*sinx*cos^4x*dx`          ...(2)
Adding(1) and (2), we get

2I = `int_0^pi x*sinx*cos^4x*dx + int_0^pi (pi - x)*sinx*cos^4x*dx`

= `int_0^pi (x + pi - x)*sinx*cos^4x*dx`

= `pi int_0^pi sinx*cos^4x*dx`

∴ I = `pi/(2) int_0^pi cos^4x*sinx*dx`

Put cos = t
∴ – sinx · dx = dt
∴ sinx · dx = – dt
When x  0, t = cos 0 = 1
When x = π cos π = – 1

∴ I = `pi/(2) int_1^(-1) t^4(- dt)`

= `- pi/(2) int_(1)^(-1) t^4*dt`

= `- pi/(2)[(t^5)/5]_1^(-1)`

= `- pi/(10)[t^5]_1^(-1)`

= `- pi/(10)[(- 1)^5 - (1)^5]`

= `- pi/(10) (- 1 - 1)`

= `(2pi)/(10)`

= `pi/(5)`.