Question

Evaluate $\int{\left( \tan x+\log \left( \sec x \right) \right)\cdot }{{e}^{x}}dx=$ \[\]
A.$\log \left( \sec x \right)+c$ \[\]
B.${{e}^{x}}\cdot \log \left( \sec x \right)+c$ \[\]
C.${{e}^{x}}\cdot \tan x+c$ \[\]
D.$-{{e}^{x}}\log \left( \sec x \right)+c$ \[\]

Answer 1

Hint: We separate the given integral using the rule of sum of indefinite integration and have$I=\int{\cdot }{{e}^{x}}\tan xdx+\int{\left( \log \left( \sec x \right) \right)}{{e}^{x}}dx$. We leave the first integral as it is and evaluate the second integral using integration by parts taking $\log \left( \sec x \right)$ as the first function and ${{e}^{x}}$as the second function. We use the formula $\dfrac{d}{dx}\log \left( \sec x \right)=\tan x$ within the evaluation. \[\]

Complete step by step answer:
We know that if there two single variable real valued integral functions say $u$ and $v$ then we integrate them by parts taking $u$ as first function and $v$ as second function using the formula
\[\int{\left( uv \right)}dx=u\int{vdx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}}dx\]
The choice of the first function depends upon how many times differentiating the function will make zero. So the rule that is used when we are integrating by parts is called ILATE, an acronym for inverse, algebraic, logarithm, trigonometric and finally exponent. It means we have to choose the first function in the order of ILATE. \[\]
We are given in the following indefinite integral to evaluate.
\[I=\int{\left( \tan x+\log \left( \sec x \right) \right)\cdot }{{e}^{x}}dx\]
We use the sum rule of indefinite integral and separate the integrand. We have,
\[I=\int{\cdot }{{e}^{x}}\tan xdx+\int{\left( \log \left( \sec x \right) \right)}{{e}^{x}}dx\]
We know the standard integral for tangnet function
\[\int{\tan xdx=}\log \left( \sec x \right)+c\]
We take differentiation both side and have
\[\begin{align}
  & \dfrac{d}{dx}\int{\tan xdx=}\dfrac{d}{dx}\left( \log \left( \sec x \right)+c \right) \\
 & \Rightarrow \tan x=\dfrac{d}{dx}\log \left( \sec x \right)+\dfrac{d}{dx}\cdot 0 \\
 & \Rightarrow \dfrac{d}{dx}\log \left( \sec x \right)=\tan x \\
\end{align}\]
We see that $\log \left( \sec x \right)$ is in the second integral$\int{\left( \log \left( \sec x \right) \right)}{{e}^{x}}dx$. So we keep the first integral $\int{\cdot }{{e}^{x}}\tan xdx$ as it is and use integration by parts for the second integral. We see that in the second integral logarithmic and exponential are in the second integral. We follow ILATE rule and choose logarithmic as the first function $u=\log \left( \sec x \right)$ and exponential $v={{e}^{x}}$ as the second function and use the formula for integration by parts. We have,
\[\begin{align}
  & I=\int{\cdot }{{e}^{x}}\tan xdx+\int{\left( \log \left( \sec x \right) \right)}{{e}^{x}}dx \\
 & =\int{\cdot }{{e}^{x}}\tan xdx+\log \left( \sec x \right)\int{{{e}^{x}}}dx-\int{\left( \dfrac{d}{dx}\log \left( \sec x \right)\int{{{e}^{x}}}dx \right)dx} \\
\end{align}\]
We use previously obtained $\dfrac{d}{dx}\log \left( \sec x \right)=\tan x$in the above step to have,
\[\begin{align}
  & =\int{\cdot }{{e}^{x}}\tan xdx+{{e}^{x}}\log \left( \sec x \right)-\int{{{e}^{x}}\tan xdx}+c \\
 & ={{e}^{x}}\log \left( \sec x \right)+c \\
\end{align}\]
Here $c$ is constant of integration.

So, the correct answer is “Option B”.

Note: We note that the given functions are smooth that are continuous and differentiable. We can directly find the integration using the formula $\int{{{e}^{x}}}\left( f\left( x \right)+{{f}^{'}}\left( x \right) \right)dx={{e}^{x}}f\left( x \right)+c$ where $f\left( x \right)$ is a smooth function. The rule of sum for two smooth functions $f\left( x \right),g\left( x \right)$ is given by $\int{\left( f\left( x \right)+g\left( x \right) \right)}=\int{f\left( x \right)}dx+\int{g\left( x \right)}dx$.