This is a classic u substitution problem. The goal of a u sub problem is to make an integral easier by substituting u in for something we do not know how to integrate.

When doing a u sub problem we want to find the "ugly" part of the problem. In this case it is the `x^2` that is inside the sin.

So we let `u = x^2`

Then we substitute in our u value into the integral and get

`int sin(u)xdx`

We are not done substituting yet though. We still need to get rid of that x, and change dx into du.

To get dx, we take the derivative of our substitution,

`u = x^2`

`du = 2xdx`

At this point we should recognize that we have xdx left in our integral, and that we have an 2xdx in the derivative of the substitution. If we move the 2 to the other side we have

`du* 1/2 = xdx`

Then we substitute 1/2 *du in for xdx in our integral and we are left with.

`int sin(u) (1/2) du`

This is much easier to integrate. Trig rules tell us the integral of sin is -cos, so our integral equals

`-(1/2)cos(u)`

But we are not done quite yet! we still have to sub x^2 back in for the u.

After doing so, our answer to this integral problem becomes

`-(1/2)cos(x^2)`

Find the integral of (sinx^2)x dx:

`int(sinu)du=-cosu+C`

So rewrite the integrand as `int(1/2sin(x^2))2xdx`

`1/2int(sinx^2)2xdx`

`"if"u=x^2,du=2xdx`

`=-1/2cos(x^2)+C`

`intsinx^2dx=?`

using change of variable(a property)

We let `x=t `

differentiating both sides we have

`2dx=dt `

`dx=dt/2 `

and replacing in above we have:

`intsint^2(dt/t)`

`1/2intsint^2dt`

the integral of sin is -cos so w'll have:

`-1/2cost^2 `

but t = x so we finally have:

`-1/2cosx^2 `

Substitite x^2 by t. So that 2x.dx=dt.

so the integral becomes 1/2 sin t dt, which is

-cos t + C, that is - cos x^2 + C.