Question

Find out the integration of the following expression: $\int{\dfrac{{{\sin }^{4}}x}{{{\cos }^{8}}x}dx}$

Answer 1

We will first start by factoring out ${{\cos }^{4}}x$ and then we will replace the term $\dfrac{{{\sin }^{4}}x}{{{\cos }^{4}}x}={{\tan }^{4}}x$ and then change $\dfrac{1}{{{\cos }^{4}}x}={{\sec }^{4}}x$ and perform integration, for that we will use the substitution method and for that we will assume $u=\tan x$ and then $du={{\sec }^{2}}xdx$ ultimately we will use power rule that is $f\left( x \right)={{x}^{n}}\Rightarrow \int{f\left( x \right)=\dfrac{{{x}^{n}}}{n+1}+C}$ and hence again replace $u=\tan x$ and get the answer.

Complete step-by-step answer:
We are given the expression: $\int{\dfrac{{{\sin }^{4}}x}{{{\cos }^{8}}x}dx}$
First will start by multiplying the expression by 1, so we will get: $\int{\dfrac{{{\sin }^{4}}x.1}{{{\cos }^{8}}x}dx}$ ,
Now from the denominator: ${{\cos }^{8}}x$ , we will factor out ${{\cos }^{4}}x$ , thus we will have the expression as: $\int{\dfrac{{{\sin }^{4}}x.1}{{{\cos }^{4}}x{{\cos }^{4}}x}dx}$
From the obtained function we will now separate the fractions that is $\int{\dfrac{1}{{{\cos }^{4}}x}.\dfrac{{{\sin }^{4}}x}{{{\cos }^{4}}x}dx}$
Now, we know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta$ , therefore $\dfrac{{{\sin }^{4}}x}{{{\cos }^{4}}x}={{\tan }^{4}}x$ and hence:
$\int{\dfrac{1}{{{\cos }^{4}}x}.\dfrac{{{\sin }^{4}}x}{{{\cos }^{4}}x}dx}\Rightarrow \int{\dfrac{1}{{{\cos }^{4}}x}.{{\tan }^{4}}xdx}\text{ }........\text{Equation 1}\text{.}$
Now, we also know that: $\dfrac{1}{\cos \theta }=\sec \theta$ , therefore $\dfrac{1}{{{\cos }^{4}}x}={{\sec }^{4}}x$ and hence, equation 1 becomes: $\int{\dfrac{1}{{{\cos }^{4}}x}.}{{\tan }^{4}}xdx=\int{{{\sec }^{4}}x.{{\tan }^{4}}xdx}$ , we will again factor out ${{\sec }^{2}}x$ from ${{\sec }^{4}}x$ and thus our integral becomes as following:
$\Rightarrow \int{{{\sec }^{4}}x.{{\tan }^{4}}xdx}=\int{\left( {{\sec }^{2}}x{{\sec }^{2}}x \right){{\tan }^{4}}xdx}\text{ }.........\text{Equation 2}$ ,
Now, according to the trigonometric identity: ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$ , therefore equation 2 becomes: $\int{\left( {{\sec }^{2}}x{{\sec }^{2}}x \right){{\tan }^{4}}xdx}=\int{\left( 1+{{\tan }^{2}}x \right){{\sec }^{2}}x{{\tan }^{4}}xdx}$ , Now, we will rearrange the expression and we will get:
$\Rightarrow \int{\left( 1+{{\tan }^{2}}x \right){{\sec }^{2}}x{{\tan }^{4}}xdx}=\int{{{\tan }^{4}}x\left( 1+{{\tan }^{2}}x \right){{\sec }^{2}}xdx}\text{ }........\text{Equation 3}$ ,
Now let $u=\tan x$ and as we know that $f\left( x \right)=\tan \theta \Rightarrow f'\left( x \right)={{\tan }^{2}}\theta$ , therefore: $du={{\sec }^{2}}xdx\Rightarrow \dfrac{1}{{{\sec }^{2}}x}du=dx$ , Now we will replace $\tan x$ as $u$ and $\dfrac{1}{{{\sec }^{2}}x}du=dx$ in equation 3, therefore:
$\Rightarrow \int{{{\tan }^{4}}x\left( 1+{{\tan }^{2}}x \right){{\sec }^{2}}xdx=\int{{{u}^{4}}\left( 1+{{u}^{2}} \right)}}{{\sec }^{2}}x.\dfrac{1}{{{\sec }^{2}}x}du$
We will now cancel: ${{\sec }^{2}}x$ , and thus we will get: $\int{{{u}^{4}}\left( 1+{{u}^{2}} \right)du}$ ,
we will now multiply ${{u}^{4}}$ into the bracket and therefore: $\int{{{u}^{4}}\left( 1+{{u}^{2}} \right)du}=\int{\left( {{u}^{4}}+{{u}^{6}} \right)du}\text{ }.........\text{Equation 4}\text{.}$
We will now use the power rule for integration that is: $f\left( x \right)={{x}^{n}}\Rightarrow \int{f\left( x \right)=\dfrac{{{x}^{n}}}{n+1}+C}$ ,
Now equation 4 will become: $\int{\left( {{u}^{6}}+{{u}^{4}} \right)du=\dfrac{{{u}^{6+1}}}{6+1}+\dfrac{{{u}^{4+1}}}{4+1}+C=\dfrac{{{u}^{7}}}{7}+\dfrac{{{u}^{5}}}{5}+C}$
Now we will again replace: $u=\tan x$ , therefore:
$\Rightarrow \dfrac{{{u}^{7}}}{7}+\dfrac{{{u}^{5}}}{5}+C=\dfrac{{{\tan }^{7}}x}{7}+\dfrac{{{\tan }^{5}}x}{5}+C$
Therefore, $\int{\dfrac{{{\sin }^{4}}x}{{{\cos }^{8}}x}dx}=\dfrac{{{\tan }^{7}}x}{7}+\dfrac{{{\tan }^{5}}x}{5}+C$

Note: We must know the basic trigonometric properties in order to find out the integral involving the trigonometric function. Some properties are known as Pythagorean identities which are the extension of Pythagoras theorem: 1). ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ , 2). ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$ , 3). $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta$ . Student might make the mistake while converting the functions and applying trigonometric functions.