Integrate: x*sin(1/2x)dx?

Question

Step by step including methods, formulas and workings plz?

Am i correct to say you use integration by parts?

Answer 1

Yes, we can use integration by parts. Actually, it seems to be the best approach here. To make things simpler, first let's integrate y * sin (y) dy (this is not necessary, but cuts down on computations). Put u = y and dv = sin(y) dy. Then, by parts,

Int u dv = uv - Int v du. v = Int sin(y) dy = - cos(y) and du = dy So,
Int y sin(y) dy = - y cos(y) - Int(- cos(y) dy = - y cos(y) + Int cos(y) dy = - y cos(y) + sin(y) + C.

Since what you realy want is Int x * sin(x/2) dx, observe that

Int x * sin(x/2) dx = 4 Int (x/2) sin(x/2) * 1/2 dx. Now, since d/dx(x/2) = 2, all we have to do is substitute x/2 for y, getting

Int x * sin(x/2) dx = 4 (-x/2)cos(x/2) + 4 sin(x/2) + C = -2x cos(x/2) + 4 sin(x/2) + C.

Answer 2

I don't think Ore...'s answer is correct. is this sin(1/(2x)) or sin(x/2)? If it's sin(1/(2x)) then I think you'll be out of luck finding a solution, so I'll assume its sin(x/2).

int(x*sin(x/2)*dx) let y = x/2 then dy = dx/2

int(2y*sin(y)*2dy) = 4*int(y*sin(y)*dy)

integration by parts, u=y dv=sin(y)dy
du = dy v = -cos(y)

4*int(y*sin(y)*dy) =4*[ -y*cos(y) - int(-cos(y)*dy)] = 4*[-y*cos(y) + sin(y)]

then int(x*sin(x/2)*dx) = -2x*cos(x/2) + 4sin(x/2)

Answer 3

xsin(x/2)dx

using turbo method:

u dv
x sin (x/2)
1 -1/2cos(x/2)
0 -1/4sin(x/2)

Then alternate signs on x, 1 and 0 and multiply diagonally and you get:
-1/2xcos(x/2)-1/4sin(x/2)

You can factor out -1/4 and get:

-1/4[2xcos(x/2)+sin(x/2)]

Answer 4

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Answer 5

= (1/2)*sin(1/(2*x))*x^2 + (1/4)*cos(1/(2*x))*x + (1/8)*sin(1/(2*x))