How can I integrate sqrt(x)*sin(sqrt(x))?

Question

Steps please

Answer 1

∫ (√x sin(√x) ) dx

To solve this, use substitution.

let z = √x. Then, if we square both sides,
z² = x, so
2z dz = dx

So our integral becomes
∫ (z sin(z) 2z dz )

Simplifying and pulling the constant 2 out of the integral,

2 ∫ (z² sin(z) dz )

At this point, we need to use integration by parts to solve this.

Let u = z². dv = sin(z) dz
du = 2z dz. v = -cos(z)

2 [ -z²cos(z) - ∫ (-cos(z)2z dz) ]

Simplifying the integral by pulling the constant (-2) out, we have

2 [ -z²cos(z) + 2 ∫ (z cos(z) dz) ]

Distributing the 2, we get

-2z²cos(z) + 4 ∫ (z cos(z) dz)

Use integration by parts again.

Let u = z. dv = cos(z) dz
du = dz. v = sin(z)

-2z²cos(z) + 4 [z sin(z) - ∫ (sin(z) dz)]

Distribute the 4,

-2z²cos(z) + 4z sin(z) - 4∫ (sin(z) dz)

And now this is easily integrable, as -cos(z).

-2z²cos(z) + 4z sin(z) - 4(-cos(z)) + C

-2z²cos(z) + 4z sin(z) + 4cos(z) + C

But z = √x, so our final answer is

-2[√x]² cos(√x) + 4√x sin(√x) + 4cos(√x) + C

Which can be simplified as

-2x cos(√x) + 4√x sin(√x) + 4cos(√x) + C

Answer 2

to integrate sqrt(x)*sin(sqrt(x))
first let t =sqrt(x)
t^2= x dx =2t dt
therefore question become

to integrate t*sin (t)*2t = 2t^2 * sin(t)
it can be solved then by multiplication rule .

Answer 3

let x = t^2
dx = 2t dt

I = int sqrtx * sin(sqrt x) dx =
int t sin(t) 2t dt =
int 2 t^2 sin(t) dt =
-2t^2 cos(t) + int 4t cos(t) =
-2t^2 cos(t) + 4t sin(t) - int 4 sin(t) =
-2t^2 cos(t) +4t sin(t) +4 cos(t) =
-2x cos(sqrt(x)) + 4sqrt(x) sin(sqrt(x)) + 4cos( sqrt(x))