Find the integral.
∫ -10 x sin x^2 dx
wouldn't that just be
5 cos(x^2) + c ?
Yes, but how would i show the work.
some patterns you should just recognize.
I know how to differentiate and integrate sin (?),
but in both cases the derivative of the "angle" which would be the x from the x^2 has to be considered.
Sure enough, the derivative of x^2, which would be 2x , is hanging around at the front as a multiple
( 5(2x) = 10x)
Formal way:
let x^2 = u
2x = du/dx
dx = du/(2x)
∫ -10 x sin x^2 dx
= ∫ -10 x sin (u) (du/(2x))
= ∫ -5 sin (u) du
= 5 cos(u) + c
= 5 cos(x^2) + c
To find the integral of ∫ -10 x sin(x^2) dx, we can use the substitution method.
Let's substitute u = x^2. Then, du/dx = 2x and dx = du/(2x).
Substituting these values into the integral, we get:
∫ -10 x sin(x^2) dx = -10 ∫ x sin(u) (du/(2x))
Simplifying, we can cancel out the x terms:
= -10/2 ∫ sin(u) du
= -5 ∫ sin(u) du
Now, we integrate the function ∫ sin(u) du:
= -5 (-cos(u)) + C
Where C is the constant of integration.
Finally, we substitute back u = x^2:
= 5 cos(x^2) + C
So, the integral of ∫ -10 x sin(x^2) dx is 5 cos(x^2) + C, where C is the constant of integration.