how do you solve the integral xsinxcosx dx [Solved]

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

Get Answer for  Who can name things that come in eights [Solved]

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)

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