INTREGAL COS2x INTEGRAL x^2 sen x ?

Question

HELP HELP

Answer 1

1. ∫ cos(2x) dx =

let 2x = u →

x = u/2 →

dx = (1/2)du

thus, substituting you get:

∫ cos(2x) dx =  Ã¢ÂˆÂ« cos u (1/2)du =

(1/2) ∫ cos u du = (1/2) sin u + C

that is, substituting back u = 2x,

∫ cos(2x) dx = (1/2) sin(2x) + C

2. ∫ x² sinx dx =

let:

sinx dx = dv → - cosx = v

x² = u → 2x dx = du

then, integrating by parts, you get:

∫ u dv = u v - ∫ v du →

∫ x² sinx dx = x²(- cosx) - ∫ (- cosx) 2x dx =

- x² cosx + 2 ∫ x cosx dx =

furtherly integrating by parts, let:

cosx dx = dv → sinx = v

x = u → dx = du

thus, integrating the remaining integral you get:

∫ u dv = u v - ∫ v du →

- x² cosx + 2 ∫ x cosx dx = - x² cosx + 2 (x sinx - ∫ sinx dx) =

- x² cosx + 2x sinx - 2 ∫ sinx dx =

- x² cosx + 2x sinx - 2(- cosx) + C =

- x² cosx + 2x sinx + 2 cosx + C

thus, in conclusion:

∫ x² sinx dx = - x² cosx + 2x sinx + 2 cosx + C

I hope it helps...

Bye!

Answer 2

The integral of cos2x is (1/2)sin2x.  This follows from the chain rule.

To integrate x^2 sin x you're looking at doing it by parts.  Choose

u=x^2 and v'=sin x

u'=2x and -v = cos x

2x cos x is going to need parts again to integrate.  Let

p=2x and q'= cox x

p'=2 and q= sin x

Hence the integral of x^2 sin x is

-x^2 cos x + 2x sin x + 2cos x