What is the integral of (sin(x^2)*x)dx Thanks!
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
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
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
Find the integral of (sinx^2)x dx:
So rewrite the integrand as `int(1/2sin(x^2))2xdx`
using change of variable(a property)
We let `x=t `
differentiating both sides we have
and replacing in above we have:
the integral of sin is -cos so w'll have:
but t = x so we finally have:
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.