$\int 2x\cdot \sin(x^2) \ dx$ = ?
My friend and I solved this question in different ways does it matter how we solve it?
I tried this way
$\int uv'= uv - \int u'v \rightarrow u=2x \ , u'=2 \ , v'=\sin(x^2) \ , v=-\cos(x^2)/2 \ ,$
$\Rightarrow \int 2x\cdot \sin(x^2)= -2x\cdot \frac{\cos(x^2)}{2}-\int2\cdot -\frac{\cos(x^2)}{2} $ $= -x\cdot \cos(x^2)+\int \cos(x^2) $ $= -x\cdot \cos(x^2)+\frac{\sin(x^2)}{2x} $
his solution was
$t=x^2 \ \rightarrow dt=2x\ dx \rightarrow dx=\frac{dt}{2x} $
$\int 2x\cdot \sin(x^2) \ dx =\int 2x\cdot \sin(t) \frac{dt}{2x}=\int \sin(t)=-\cos(x^2) $
His method is actually much simpler than integration by parts. He did it by simple substitution with $t = x^2, dt = 2x \ dx$ to get $\int \sin t \ dt = - \cos t + C = -\cos(x^2) +C.$
You want to find the simplest method in integrating a function, but unfortunately, integrating this function by parts is a nightmare (and it didn't result in the original function when it's differentiated).