Integral of $$$e^{\frac{\pi t}{2}}$$$

Find $$$\int e^{\frac{\pi t}{2}}\, dt$$$.

Let $$$u=\frac{\pi t}{2}$$$.

Then $$$du=\left(\frac{\pi t}{2}\right)^{\prime }dt = \frac{\pi}{2} dt$$$ (steps can be seen »), and we have that $$$dt = \frac{2 du}{\pi}$$$.

Thus,

$${\color{red}{\int{e^{\frac{\pi t}{2}} d t}}} = {\color{red}{\int{\frac{2 e^{u}}{\pi} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{2}{\pi}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\frac{2 e^{u}}{\pi} d u}}} = {\color{red}{\left(\frac{2 \int{e^{u} d u}}{\pi}\right)}}$$

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

$$\frac{2 {\color{red}{\int{e^{u} d u}}}}{\pi} = \frac{2 {\color{red}{e^{u}}}}{\pi}$$

Recall that $$$u=\frac{\pi t}{2}$$$:

$$\frac{2 e^{{\color{red}{u}}}}{\pi} = \frac{2 e^{{\color{red}{\left(\frac{\pi t}{2}\right)}}}}{\pi}$$

Therefore,

$$\int{e^{\frac{\pi t}{2}} d t} = \frac{2 e^{\frac{\pi t}{2}}}{\pi}$$

Add the constant of integration:

$$\int{e^{\frac{\pi t}{2}} d t} = \frac{2 e^{\frac{\pi t}{2}}}{\pi}+C$$

$$$\int e^{\frac{\pi t}{2}}\, dt = \frac{2 e^{\frac{\pi t}{2}}}{\pi} + C$$$A