Integrate the given trigonometric expression.

\[\int{\dfrac{{{e}^{x}}.\cos 3x-{{4}^{x}}\tan 3x}{{{4}^{x}}.\cos 3x}}.dx\]

Hint: Get two fractions in subtraction, with dividing the numerator terms individually. Use property of surd \[{{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}.\] Integration of\[{{a}^{x}}\], is given as\[\int{{{a}^{x}}dx=}\dfrac{{{a}^{x}}}{{{\log }_{e}}a}\]. Differentiation of \[\sec x\] is given as\[\sec x\tan x\], use this concept while using the substitution approach with the integration of the second term formed after separating the given expression in two individual terms.

Complete step-by-step solution -
Given integral in the question is
\[\int{\dfrac{{{e}^{x}}.\cos 3x-{{4}^{x}}\tan 3x}{{{4}^{x}}.\cos 3x}}.dx\]\[\to (1)\]
Let us divide \[{{e}^{x}}\cos 3x-{{4}^{x}}\tan 3x\] by\[{{4}^{x}}\cos 3x\], hence, we get
\[\begin{align}
  & =\int{\left( \dfrac{{{e}^{x}}\cos 3x}{{{4}^{x}}\cos 3x}-\dfrac{{{4}^{x}}\tan 3x}{{{4}^{x}}\cos 3x} \right)}dx \\
 & =\int{\left( \dfrac{{{e}^{x}}}{{{4}^{x}}}-\dfrac{\tan 3x}{\cos 3x} \right)dx} \\
\end{align}\]
We know that property of surds, given as
\[\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}\]
So, we can get integral with the help of above relation as
\[\begin{align}
  & =\int{\left( {{\left( \dfrac{e}{4} \right)}^{x}}-\dfrac{\tan 3x}{\cos 3x} \right)dx} \\
 & =\int{{{\left( \dfrac{e}{4} \right)}^{x}}dx-\int{\dfrac{\tan 3x}{\cos 3x}dx}}\to (2) \\
\end{align}\]
Now, let us suppose the \[\int{{{\left( \dfrac{e}{4} \right)}^{x}}dx}\]as\[{{\text{I}}_{1}}\], and \[\int{\dfrac{\tan 3x}{\cos 3x}dx}\]as\[{{\text{I}}_{\text{2}}}\]. So, let us calculate \[{{\text{I}}_{1}}\]and \[{{\text{I}}_{\text{2}}}\]one by one. So, we have
\[{{\text{I}}_{1}}=\int{{{\left( \dfrac{e}{4} \right)}^{x}}dx}\]\[\to (3)\]
\[{{\text{I}}_{\text{2}}}=\int{\dfrac{\tan 3x}{\cos 3x}dx}\]\[\to (4)\]
So, let us calculate\[{{\text{I}}_{1}}\]as
\[{{\text{I}}_{1}}=\int{{{\left( \dfrac{e}{4} \right)}^{x}}dx}\]
We know the integration of \[{{a}^{x}}\]is given as
\[\int{{{a}^{x}}}dx=\dfrac{{{a}^{x}}}{{{\log }_{e}}a}\]
Hence, value of \[{{\text{I}}_{1}}\]is given as
\[{{\text{I}}_{1}}=\dfrac{{{\left( \dfrac{e}{4} \right)}^{x}}}{{{\log }_{e}}\left( \dfrac{e}{4} \right)}+c\]
We know that the property of logarithm function is given as
\[{{\log }_{c}}\left( \dfrac{a}{b} \right)={{\log }_{c}}a-{{\log }_{c}}b\]
So, we get \[{{\text{I}}_{1}}\]as
\[{{\text{I}}_{1}}={{\left( \dfrac{e}{4} \right)}^{x}}\dfrac{1}{{{\log }_{e}}e-{{\log }_{e}}4}+c\]
We know \[{{\log }_{a}}a=1\]. So, we get value of \[{{\text{I}}_{1}}\]as
\[{{\text{I}}_{1}}=\dfrac{{{e}^{x}}}{{{4}^{x}}\left( 1-{{\log }_{e}}4 \right)}+c\]\[\to (5)\]
Now, we can evaluate value of \[{{\text{I}}_{\text{2}}}\]as
\[{{\text{I}}_{\text{2}}}=\int{\dfrac{\tan 3x}{\cos 3x}}dx\]\[\to (6)\]
Now, we know the relation between \[\cos \theta \]and \[\sec \theta \]is given as
\[\sec \theta =\dfrac{1}{\cos \theta }\]
Hence, integral \[{{\text{I}}_{\text{2}}}\]can be written as
\[{{\text{I}}_{\text{2}}}=\int{\sec 3x\tan 3x\text{ }dx}\]\[\to (7)\]
Now, suppose \[\sec 3x=t\]. Differentiate the above relation with respect to\['x'\]. We get
\[\dfrac{d}{dx}\left( \sec 3x \right)=\dfrac{dt}{dx}\]
We know that the derivative of \[\sec \theta \]is\[\sec \theta \tan \theta \]. Hence, we get
\[\begin{align}
  & \dfrac{d}{dx}\left( \sec 3x \right)=\dfrac{dt}{dx} \\
 & 3\sec 3x\tan 3x=\dfrac{dt}{dx} \\
\end{align}\]
\[\sec 3x\tan 3x\text{ }dx=\dfrac{dt}{3}\]\[\to (8)\]
Now, replacing \[\sec 3x\tan 3x\text{ }dx\]by \[\dfrac{dt}{3}\] in the expression\[{{\text{I}}_{\text{2}}}\]. So, we get
\[{{\text{I}}_{\text{2}}}=\int{\dfrac{dt}{3}}=\dfrac{t}{3}+c\]
Now, put \[t=\sec 3x\]to the above equation. So, we get
\[{{\text{I}}_{\text{2}}}=\dfrac{\sec 3x}{3}+c\]\[\to (9)\]
Hence, we get the integral of the given expression in the problem from the equation (2), (7) and (9) as
\[={{\left( \dfrac{e}{4} \right)}^{x}}\dfrac{1}{\left( 1-{{\log }_{e}}4 \right)}-\dfrac{\sec 3x}{3}+c\]
So, we get
\[\int{\dfrac{{{e}^{x}}\cos 3x-{{4}^{x}}\tan 3x}{{{4}^{x}}\cos 3x}}dx={{\left( \dfrac{e}{4} \right)}^{x}}\dfrac{1}{\left( 1-{{\log }_{e}}4 \right)}-\dfrac{\sec 3x}{3}+c\]

Note: One may get confused with the given term integral, as terms of exponential and trigonometric both are involved. So, one may think of applying integration by parts, which is a complex approach. Just find two fractions by dividing the numerators (in difference) by denominator individually.
Students make a lot of mistakes with the results of \[\int{{{a}^{x}}}dx\] and \[\dfrac{d}{dx}{{a}^{x}}\]. So, both are given as \[\int{{{a}^{x}}}dx=\dfrac{{{a}^{x}}}{{{\log }_{e}}a}\] and \[\dfrac{d}{dx}{{a}^{x}}={{a}^{x}}{{\log }_{e}}a\]. So, one may get confused with them.