Question

Solve the given question in a detail manner:
Integrate the following function:
\[f\left( x \right) = \dfrac{1}{{8{{\cos }^2}x + 3{{\sin }^2}x + 1}}\]

Answer 1

Hint: Take the given expression in the form of integral solving. Divide the numerator and the denominator with \[{\cos ^2}x\] and simplify the terms accordingly. Then take the parameter for the equation. Differentiate the parameter on the both sides with respect to the variable. Then substitute it in the integral equation. Solve the equation and finally substitute the value of the parameter back in the integral equation to get the final solution.

Complete step-by-step solution:
Given function;
\[f\left( x \right) = \dfrac{1}{{8{{\cos }^2}x + 3{{\sin }^2}x + 1}}\]
We take the function as \[I\]
We divide the numerator and the denominator with \[{\cos ^2}x\]
\[ \Rightarrow I = \int {\dfrac{{\dfrac{1}{{{{\cos }^2}x}}}}{{\dfrac{{8{{\cos }^2}x}}{{{{\cos }^2}x}} + \dfrac{{3{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{1}{{{{\cos }^2}x}}}}} dx\]
We can write these terms as given below;
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x}}{{8 + 3{{\tan }^2}x + {{\sec }^2}x}}dx} \]
We can write \[{\sec ^2}x = 1 + {\tan ^2}x\] using the trigonometric functions;
Replacing\[{\sec ^2}x = 1 + {\tan ^2}x\], we get;
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x}}{{8 + 3{{\tan }^2}x + \left( {1 + {{\tan }^2}x} \right)}}dx} \]
We simplify the denominator of the above equation by adding the like terms
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x}}{{4{{\tan }^2}x + 9}}dx} \]
We can write the constant term outside of the integral. Then we get;
\[ \Rightarrow I = \dfrac{1}{4}\int {\dfrac{{{{\sec }^2}x}}{{{{\tan }^2}x + \dfrac{9}{4}}}dx} \]
Let us take a parameter to the equation so that we can simplify the equation.
Let us consider,
\[t = \tan x\]
Differentiating both the sides with respect to \[x\], we get;
\[\dfrac{{dt}}{{dx}} = {\sec ^2}x\]
Now, taking the denominator to the other side, we get;
\[ \Rightarrow dt = {\sec ^2}xdx\]
We have the numerator of the integral, i.e.,
\[ \Rightarrow I = \dfrac{1}{4}\int {\dfrac{{{{\sec }^2}x}}{{{{\tan }^2}x + \dfrac{9}{4}}}dx} \]
Here, we can substitute the numerator with,
\[ \Rightarrow dt = {\sec ^2}xdx\]
We get;
\[ \Rightarrow I = \dfrac{1}{4}\int {\dfrac{{dt}}{{{{\tan }^2}x + \dfrac{9}{4}}}} \]
We can write the denominator as the perfect squares and replacing the first term of the denominator with the parameter taken. Then, we get;
\[ \Rightarrow I = \dfrac{1}{4}\int {\dfrac{{dt}}{{{\operatorname{t} ^2} + {{\left( {\dfrac{3}{2}} \right)}^2}}}} \]
Now, this is in the form of the integral \[\int {\dfrac{{dx}}{{{x^2} + {a^2}}}} \];
This can be expanded as given below;
\[\int {\dfrac{{dx}}{{{x^2} + {a^2}}}} = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right) + c\]
Comparing the obtained equation with this equation, we have;
\[x = t\]
\[a = \dfrac{3}{2}\]
Substituting the values, we get;
\[ \Rightarrow I = \dfrac{1}{4} \times \dfrac{1}{{\dfrac{3}{2}}}{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{t}{1}}}{{\dfrac{3}{2}}}} \right) + c\]
Simplifying the terms, we get;
\[ \Rightarrow I = \dfrac{1}{6}{\tan ^{ - 1}}\left( {\dfrac{{2t}}{3}} \right) + c\]
Now substituting the value of the parameter back, we get;
\[ \Rightarrow I = \dfrac{1}{6}{\tan ^{ - 1}}\left( {\dfrac{{2\tan x}}{3}} \right) + c\]

Therefore, we have the integral value;
\[f\left( x \right) = \dfrac{1}{{8{{\cos }^2}x + 3{{\sin }^2}x + 1}} = \dfrac{1}{6}{\tan ^{ - 1}}\left( {\dfrac{{2\tan x}}{3}} \right) + c\]

Note: We can derive that \[{\sec ^2}x = 1 + {\tan ^2}x\].
Let us take the known equation, \[{\sin ^2}x + {\cos ^2}x = 1\]
By dividing each term with \[{\cos ^2}x\], we get;
\[\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{{{\cos }^2}x}}{{{{\cos }^2}x}} = \dfrac{1}{{{{\cos }^2}x}}\]
Now, we can write these values as
\[{\tan ^2}x + 1 = {\sec ^2}x\]