Question

Integrate the given trigonometric expression.

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

Answer 1

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

& =\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{aligned}$$ We know that property of surds, given as $$\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$$ So, we can get integral with the help of above relation as $$\begin{aligned} & =\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{aligned}$$ Now, let us suppose the $$\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 $$\begin{aligned} & \dfrac{d}{dx}\left( \sec 3x \right)=\dfrac{dt}{dx} \\\ & 3\sec 3x\tan 3x=\dfrac{dt}{dx} \\\ \end{aligned}$$ $$\sec 3x\tan 3x\text{ }dx=\dfrac{dt}{3}$$$$\to (8)$$ 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.