I am trying to define the steps for the integration of the following $$\int\limits_{0}^{\frac{\pi}{2}}\left((\cos(t))(\cos^2(t)+2\sin(t)e^{\cos(t)}+1)+(-\sin(t))(2\sin(t)\cos(t)+\sin^2(t)e^{\cos(t)}+2\cos(t))\right)dt$$

I am plugging this into wolfram alpha and it shows that the answer is 1, but I have no idea how to get there (the step-by-step solution does not work for problems of this size).


It looks like you are trying to calculate $$ \int_\gamma P\,dx+Q\,dy=\int_\gamma(2xy+y^2e^x+2x)\,dx+(x^2+2ye^x+1)\,dy $$ where $\gamma$ is a quarter circle of radius one, starting in $(1,0)$ ending in $(0,1)$.

Now, since it happens that $$ \frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y} $$ you will actually have a potential function, which simplifies the calculations very much. Can you proceed by this hint, or do you need further help?


The next step is to find a potential function, i.e. a function $U$ that satisfies $$ \frac{\partial U}{\partial x}=P,\quad\text{and}\quad \frac{\partial U}{\partial y}=Q. $$ In this case, one can almost find one by staring. Your book surely gives a method on how to do it in general. Move mouse over box below to find one potential.


Once that is done, your integral is calculated by $$ \int_\gamma(2xy+y^2e^x+2x)\,dx+(x^2+2ye^x+1)\,dy=U(0,1)-U(1,0). $$

  • $\begingroup$ I realized this relationship earlier, but I wasn't sure as to how to apply it in integration. what are the next steps? $\endgroup$ – user2615936 Dec 1 '15 at 8:46
  • $\begingroup$ I gave more steps... I suggest that you try to find $U$ by the methods explained in the book... It is not always one can see the potential direclty like this. $\endgroup$ – mickep Dec 1 '15 at 8:51

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