Question

Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`

Answer 1

Let I = `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`

= `int_0^π sin^2x(1 + 2 cosx)(1 + cosx)^2.sinx.dx`

= `int_0^π (1 - cos^2x)(1 +  2cosx)(1 + cosx)^2.sinx.dx`

Put cos x = t

∴ – sinx.dx = dt.

∴ sinx.dx = –dt

When x = 0, t = cos 0 = 1

When x = π, t = cos π = –1

∴ I = `int_1^(-1) (1 - t^2)(1 + 2t)(1 + t)^2(- dt)`

= `-int_1^(-1)(1 + 2t - t^2 - 2t^3)(1 + 2t + t^2).dt`

= `- int_1^(-1) (1 + 2t - t^2 - 2t^3 + 2t + 4t^2 - 2t^3 - 4t^4 + t^2 + 2t^3 - t^4 - 2t^5).dt`

= `int_1^(-1) (1 + 4t + 4t^2 - 2t^3 - 5t^4 - 2t^5).dt`

= `-[t + 4(t^2/2) + 4(t^3/3) - 2(t^4/4) - 5(t^5/5) - 2(t^6/6)]_1^(-1)`

= `-[t + 2t^2  4/3t^3 - 1/2t^4 - t^5 - 1/3t^6]_1^(-1)`

= `-[(-1 + 2 - 4/3 - 1/2 + 1 - 1/3) - (1 + 2 + 4/3 - 1/2 - 1 - 1/3)]`

= `-[-1 + 2 - 4/3 - 1/2 + 1 - 1/3 - 1 - 2 - 4/3 + 1/2 + 1 + 1/3]`

= `-[-8/3]`

= `(8)/(3)`.