Question

Evaluate the Integral $\int {\mathrm{e}}^{\mathrm{x}}\left[{\sec }^2\left(\mathrm{x}\right)+\hspace{0.17em}\tan \left(\mathrm{x}\right)\right]\mathrm{d}x$.

Answer 1

Finding $\int {\mathrm{e}}^{\mathrm{x}}\left[{\sec }^2\left(\mathrm{x}\right)+\hspace{0.17em}\tan \left(\mathrm{x}\right)\right]\mathrm{d}x$:

$\begin{array}{rcl}\int {\mathrm{e}}^x\left[{\sec }^2\left(x\right)+\tan \left(x\right)\right]& =& \int \left[{\mathrm{e}}^{\mathrm{x}}{\sec }^2\left(\mathrm{x}\right)+{\mathrm{e}}^{\mathrm{x}}\tan \left(\mathrm{x}\right)\right]\mathrm{d}x\\ & =& \int {\mathrm{e}}^{\mathrm{x}}{\sec }^2\left(\mathrm{x}\right)\mathrm{d}x+\int {\mathrm{e}}^{\mathrm{x}}\tan \left(\mathrm{x}\right)\mathrm{d}x\end{array}$

Integration by parts is done when conventional methods of integration do not work on the integrand.

The formula for integration by parts is given as,

$\int f\left(x\right)·g\left(x\right)\mathrm{d}x=f\left(x\right)\int g\left(x\right)dx-\int \left(f\text{'}\left(x\right)\int g\left(x\right)\mathrm{d}x\right)\mathrm{d}x$

[General tip: When picking the functions to be substituted into the formula, $f\left(x\right)$ must be a function that is easily differentiable or goes to zero when differentiated repeatedly. Likewise, $g\left(x\right)$ must be the function that is more easily integrated]

Let, $f\left(x\right)={\mathrm{e}}^x$ and $g\left(x\right)=\sec \left(x\right)$ in the first integrand. Thus,

$\begin{array}{rcl}\int {\mathrm{e}}^x{\sec }^2\left(x\right)\mathrm{d}x+\int {\mathrm{e}}^x\tan \left(x\right)\mathrm{d}x& =& {\mathrm{e}}^x\int {\sec }^2\left(x\right)\mathrm{d}x-\int \left[\frac{d}{d\mathrm{x}}{\mathrm{e}}^x\int {\sec }^2\left(x\right)dx\right]\mathrm{d}x+\int {\mathrm{e}}^{\mathrm{x}}\tan \left(x\right)\mathrm{d}x+\mathrm{C}\\ & =& {\mathrm{e}}^x\tan \left(x\right)-\int {\mathrm{e}}^x\tan \left(x\right)\mathrm{d}x+\int {\mathrm{e}}^{\mathrm{x}}\tan \left(\mathrm{x}\right)\mathrm{d}x+\mathrm{C}\left[\because \int {\sec }^2\left(\mathrm{x}\right)=\tan \left(\mathrm{x}\right)\right]\\ & =& {\mathrm{e}}^x\tan \left(x\right)+\mathrm{C}\end{array}$

Hence, $\int {\mathrm{e}}^{\mathrm{x}}\left[{\sec }^2\left(\mathrm{x}\right)+\hspace{0.17em}\tan \left(\mathrm{x}\right)\right]={\mathrm{e}}^x\tan \left(\mathrm{x}\right)+\mathrm{C}$.