# Finding the value of $\int{F \cdot dr}$

Find the value of $$\int{F \cdot \mathrm{d}r}$$, where $$F(x,y) = \langle 5e^y+ye^x,e^x+5xe^y \rangle$$ and $$C: r(t) = \left\langle\sin\left(\frac{\pi t}{2}\right),\ln(t)\right\rangle; 1\le t\le2$$ So far, I've tried substituting in using the parametriziation suggested by the $$r(t)$$ function into the $$F(x,y)$$ vector field but that produced such a big integral there's no way of evaluating it. Any help would be welcome, thank you in advance!

• If \begin{align} r &= \langle x(t), y(t) \rangle \\ &= \langle \sin(at), \ln(t) \rangle \end{align} then \begin{align} dr &= \langle dx, dy \rangle \\ &= \langle a \cos(at) dt, dt/t \rangle \end{align} and \begin{align} F &= \langle 5e^y + ye^x, e^x + 5xe^x \rangle \\ &= \langle 5t + \ln(t) e^{\sin(at)}, e^{\sin(at)} + 5 \sin(at) e^{\sin(at)} \rangle \end{align} and dotting gives $$\int_{1}^{2} [5t + \ln(t) e^{\sin(at)}] a \cos(at) + \frac{e^{\sin(at)} + 5 \sin(at) e^{\sin(at)}}{t} dt$$ To solve, use integration by parts and $u$ substitutions. – Mattos Mar 21 at 4:06
• Well, for one, integration by parts on $\ln(t) \cdot a \cos(at) e^{\sin(at)}$, with $u = \ln(t) \implies u' = 1/t$ and $v' = a \cos(at) e^{\sin(at)} \implies v = e^{\sin(at)}$ gives $$I = \ln(t) e^{\sin(at)} \bigg \lvert_{1}^{2} - \int_{1}^{2} \frac{e^{\sin(at)}}{t} dt$$ where the integral part of $I$ cancels with the $$\color{red} + \int_{1}^{2} \frac{e^{\sin(at)}}{t} dt$$ from our original integral. – Mattos Mar 21 at 4:18
• @Mattos This should be an answer, not a comment. – Naman Kumar Mar 21 at 4:52

As seen in the comments (and corrected on one key point), we can parametrize and set up the integral $$\int_C F\cdot d\mathbf{r} = \int_1^2 [5t+\ln(t)e^{\sin(at)}]a\cos at + \frac{e^{\sin(at)}+5\sin(at)e^{\ln t}}{t}\,dt$$ where $$a=\frac{\pi}{2}$$, abbreviated to simplify the typography.
Instead, note that we can write $$F$$ as a gradient $$F=\nabla G$$, where $$G(x,y)=5xe^y+ye^x$$. Then $$G$$ acts as an antiderivative; we can apply the line integral version of the FTC to get $$\int_C F\cdot d\mathbf{r} = G(r(2))-G(r(1)) = G(0,\ln 2) - G(1,0) = \ln 2 - 5$$