# Stefan's question about a line integral.

#### Prove It

##### Well-known member
MHB Math Helper
Evaluate the line integral

\displaystyle \begin{align*} \int_{(0,0,0)}^{ \left( \frac{1}{3}, \frac{\pi}{2}, 1 \right) }{ \left[ 12\,z^2 + 4\,\mathrm{e}^{4x}\cos{(5y)} \right] \,\mathrm{d}x - \left[ 5\,\mathrm{e}^{4x}\sin{(5y)} \right] \,\mathrm{d}y + 24\,x\,z\,\mathrm{d}z } \end{align*}
The line starting at \displaystyle \begin{align*} \left( 0, 0, 0 \right) \end{align*} and ending at \displaystyle \begin{align*} \left( \frac{1}{3}, \frac{\pi}{2}, 1 \right) \end{align*} can be expressed in a parametric (vector) form as \displaystyle \begin{align*} \left( x, y, z \right) = \left( \frac{1}{3}t , \frac{\pi}{2}t, t \right) \end{align*} with \displaystyle \begin{align*} 0 \leq t \leq 1 \end{align*}. Thus

\displaystyle \begin{align*} x &= \frac{1}{3}t \implies \mathrm{d}x = \frac{1}{3}\mathrm{d}t \\ y &= \frac{\pi}{2}t \implies \mathrm{d}y = \frac{\pi}{2}\mathrm{d}t \\ z &= t \implies \mathrm{d}z = \mathrm{d}t \end{align*}

and so the line integral becomes

\displaystyle \begin{align*} &= \int_0^1{ \left\{ 12t^2 + 4\mathrm{e}^{ 4 \left( \frac{1}{3} t \right) } \cos{ \left[ 5 \left( \frac{\pi}{2}t \right) \right] } \right\} \frac{1}{3}\mathrm{d}t - \left\{ 5\mathrm{e}^{4 \left( \frac{1}{3}t \right) } \sin{ \left[ 5 \left( \frac{\pi}{2}t \right) \right] } \right\} \frac{\pi}{2}\mathrm{d}t + 24 \left( \frac{1}{3}t \right) t \, \mathrm{d}t } \\ &= \int_0^1{ 4\,t^2 + \frac{4}{3}\,\mathrm{e}^{ \frac{4}{3}t } \cos{ \left( \frac{5\pi}{2}t \right)} - \frac{5\pi}{2}\,\mathrm{e}^{\frac{4}{3}t}\sin{ \left( \frac{5\pi}{2}t \right) } + 8\,t^2 \, \mathrm{d}t } \\ &= \int_0^1{ 12\,t^2 + \frac{4}{3}\,\mathrm{e}^{\frac{4}{3}t}\cos{ \left( \frac{5\pi}{2}t \right) } - \frac{5\pi}{2}\,\mathrm{e}^{\frac{4}{3}t} \sin{ \left( \frac{5\pi}{2} t\right) } \, \mathrm{d}t } \\ \end{align*}

Now to integrate these, the product functions either require integration by parts, or use of the rules \displaystyle \begin{align*} \int{ \mathrm{e}^{b\,x}\sin{(a\,x)} \, \mathrm{d}x } = \frac{1}{a^2 + b^2} \, \mathrm{e}^{b\,x} \, \left[ b\sin{(a\,x)} - a\cos{(a\,x)} \right] + C \end{align*} and \displaystyle \begin{align*} \int{ \mathrm{e}^{b\,x}\cos{(a\,x)} \,\mathrm{d}x } = \frac{1}{a^2 + b^2}\,\mathrm{e}^{b\,x} \, \left[ a\sin{(a\,x)} + b\cos{(a\,x)} \right] + C \end{align*}.