x sinx cosx dx =

note that sinx cosx = (1/2) sin 2x

(1/2) x sin 2x dx =

now let x = u; sin 2x dx = dv

dx = du; (-1/2)cos 2x = v

then, proceeding by parts,

(1/2) x sin 2x dx =

(1/2) [(-1/2)x cos 2x (-1/2)cos 2x dx] =

(1/2) [(-1/2)x cos 2x (1/2) cos 2x dx] =

(-1/4)x cos 2x (1/4) cos 2x dx =

(-1/4)x cos 2x (1/4)(1/2) sin 2x c =

(-1/4)x cos 2x (1/8) sin 2x c =

(expressing it again in terms of sinx, cosx)

(-1/4)x (cosx sinx) (1/8) 2sinx cosx c =

(-1/4)x cosx (1/4)x sinx (1/4)sinx cosx c

(Ive just successfully checked the result)

You will use integration by parts to do this one.

we have , sin 2x = 2 sinx cos x , sinx cosx = sin 2x / 2

therefore

integral ( x sinx cosx ) dx

= integral ( x sin 2x / 2 ) dx

let , u = x , du = dx

dv = (sin 2x / 2 )dx , v = cos 2x / 4 c

= ( 1 / 4) x cos 2x ) ( 1 / 4) integral ( cos 2x ) dx

= ( 1 / 4) x cos 2x ( 1 / 4) integral cos 2x dx

= ( 1 / 4) x cos 2x ( 1 / 8 ) sin 2x c

use integration by parts to start, with

u = x

du = dx

dv = sin(x)cos(x) dx

to integrate sin(x)cos(x)dx let w = sin(x), so dw = cos(x)dx

so we get wdw which is w/2 eliminating w gives sin(x)/2

u = x

dv = sin(x) cos(x) dx

du = dx

v = [sin(x)]/2

applying the by parts fromual:

xsin(x)/2 sin(x)/2 dx

apply the power reduction formula:

sin(x) = (1 cos(2x)) / 2

xsin(x)/2 1/41 cos(2x) dx

which integrates to

xsin(x)/2 1/4{x (1/2)sin(2x)) C

which cleans up to

(x/2)sin(x) x/4 sin(2x)/8 C

applying the by parts fromual:

xsin(x)/2 sin(x)/2 dx

apply the power reduction formula:

sin(x) = (1 cos(2x)) / 2

xsin(x)/2 1/41 cos(2x) dx

which integrates to

xsin(x)/2 1/4{x (1/2)sin(2x)) C

(x/2)(sin(x) ) sin(x)cos(x)/4 C

dividing and multiply eq. by 2

so, 2xsinxcosx dx / 2

now, take 2sinxcosx = sin2x

so, xsin2x dx / 2

now taking integration by parts

1/2 (x (integ sin2x dx) integ ((x dx)(integ sin2x dx)))

(xcos2x /4 ) integ (cos2x/4 dx)

(xcos 2x /4) (sin2x / 8)

is the answer

uv v du

= x*[sin(x)]/2 [sin(x)]/2 dx

= x*[sin(x)]/2 (1/2) sin(x) dx

Trig identity:

= x*[sin(x)]/2 (1/2) [1 cos(2x)]/2 dx

= x*[sin(x)]/2 (1/4) [1 cos(2x)] dx

= x*[sin(x)]/2 (1/4)[x (1/2)sin(2x)] C

= x*[sin(x)]/2 (1/4)x (1/8)sin(2x) C

x sin(x) cos(x) dx

u = x

dv = sin(x) cos(x) dx

du = dx

v = [sin(x)]/2

uv v du

= x*[sin(x)]/2 [sin(x)]/2 dx

= x*[sin(x)]/2 (1/2) sin(x) dx

Trig identity:

= x*[sin(x)]/2 (1/2) [1 cos(2x)]/2 dx

= x*[sin(x)]/2 (1/4) [1 cos(2x)] dx

= x*[sin(x)]/2 (1/4)[x (1/2)sin(2x)] C

= x*[sin(x)]/2 (1/4)x (1/8)sin(2x) C

This post is last updated on hrtanswers.com at Date : 1st of September – 2022