I'm doing a course in Complex analysis and found myself struggling with a specific line integral.
Let $$\gamma (t) \equiv \pi t + (t^2-t)i,\qquad 0 \leq t \leq 1$$
Compute the line integral
$$ \int z \sin z dz$$ over the line $\gamma$.
I know that $dz=(\pi+i(2t-1))dt$, and could insert this and substitute $z=\gamma(t)$. Limits on the integral run from $t=0$ to $t=1$. Then I would have: $$\int_0^1(\pi t +i(t^2-t))(\sin(\pi t+i(t^2-t))(\pi+i(2t-1))dt$$
I could re-write this in the form $\int u \sin (u) u'$ using integration by parts but what do I do next?
Since the curve you are working on is quite difficult to handle, you could start thinking of the Fundamental Theorem. Indeed, the function $f(z)=\sin z-z\cos z$ is an antiderivative of $z\sin z$ and the curve gamma connects the points $z=0,~z=\pi$, so: $$\int_\gamma z\sin zdz=\int_\gamma f'(z)dz=f(\pi)-f(0)=\pi.$$