Question

Evaluate the following integral
$\int{\dfrac{{{\operatorname{Sin}}^{6}}x+{{\operatorname{Cos}}^{6}}x}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}dx}$

Answer 1

Hint: Write the term in the numerator ${{\operatorname{Sin}}^{6}}x+{{\operatorname{Cos}}^{6}}x$ as ${{\left( {{\operatorname{Sin}}^{2}}x \right)}^{3}}+{{\left( {{\operatorname{Cos}}^{2}}x \right)}^{3}}$, we can see that it became in the form of ${{a}^{3}}+{{b}^{3}}$ and then apply the formula of ${{a}^{3}}+{{b}^{3}}$ i.e. ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$ . Simply it further and then integrate the resulting terms separately using the basic formulae of integration.

Complete step by step answer:
The given integrand seems complicated and it can not be integrated directly.
So, first we have to simplify it.
Consider, the given integral,
$\int{\dfrac{{{\operatorname{Sin}}^{6}}x+{{\operatorname{Cos}}^{6}}x}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}dx}$
We will first write the numerator of the integrand as ${{\left( {{\operatorname{Sin}}^{2}}x \right)}^{3}}+{{\left( {{\operatorname{Cos}}^{2}}x \right)}^{3}}$ and get the integral in form,
$\int{\dfrac{{{\left( {{\operatorname{Sin}}^{2}}x \right)}^{3}}+{{\left( {{\operatorname{Cos}}^{2}}x \right)}^{3}}}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}dx}$
Now, we can see that the numerator of the integrand is in the form, ${{a}^{3}}+{{b}^{3}}$
We know that the formula of ${{a}^{3}}+{{b}^{3}}$ is ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
Now, applying this formula to the numerator of integrand , we get
$\int{\dfrac{\left( {{\operatorname{Sin}}^{2}}x+{{\operatorname{Cos}}^{2}}x \right)\left( {{\operatorname{Sin}}^{4}}x+{{\operatorname{Cos}}^{4}}x-{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x \right)}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}dx}$
We know from the trigonometric identities that
${{\operatorname{Sin}}^{2}}x+{{\operatorname{Cos}}^{2}}x=1$
So, now the integral becomes,
$\int{\dfrac{\left( {{\operatorname{Sin}}^{4}}x+{{\operatorname{Cos}}^{4}}x-{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x \right)}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}dx}$
Now, adding and subtracting $2{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x$ to the numerator of the integrand, we get
$\int{\dfrac{\left( {{\operatorname{Sin}}^{4}}x+{{\operatorname{Cos}}^{4}}x+2{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x-3{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x \right)}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}dx}$
Now, If you see the first 3 terms of numerator, it is an identity
${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$
So, using this identity the integral becomes,
$\int{\dfrac{\left( {{\left( {{\operatorname{Sin}}^{2}}x+{{\operatorname{Cos}}^{2}}x \right)}^{2}}-3{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x \right)}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}dx}$
Again using the trigonometric identity, we get
$\int{\dfrac{1-3{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}dx}$
Now, separating the terms in the integrand, we get
$\int{\left( \dfrac{1}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}-3 \right)dx}$
$\int{\left( \dfrac{1}{{{\operatorname{Cos}}^{2}}x}+\dfrac{1}{{{\operatorname{Sin}}^{2}}x}-3 \right)dx}$
We know from trigonometric identities that,
$\dfrac{1}{{{\operatorname{Cos}}^{2}}x}={{\operatorname{Sec}}^{2}}x$ and $\dfrac{1}{{{\operatorname{Sin}}^{2}}x}={{\operatorname{Cosec}}^{2}}x$
Using these identities in the last step, we get
$\int{\left( {{\operatorname{Sec}}^{2}}x+{{\operatorname{Cosec}}^{2}}x-3 \right)dx}$
Now, we can use the basic integration formulas,
$\int{{{\operatorname{Sec}}^{2}}xdx=}\operatorname{Tan}x+C$ , $\int{{{\operatorname{Cosec}}^{2}}xdx=-\operatorname{Cot}x}+C$
And get the value of integral
$=\operatorname{Tan}x-\operatorname{Cot}x-3x+C$

Hence,
$\int{\dfrac{{{\operatorname{Sin}}^{6}}x+{{\operatorname{Cos}}^{6}}x}{{{\operatorname{Sin}}^{2}}x{{\operatorname{Cos}}^{2}}x}dx}$ $=\operatorname{Tan}x-\operatorname{Cot}x-3x+C$

Note: The alternate method to solve this integral is by separating the integrand into 2 terms i.e.
\[\dfrac{{{\operatorname{Sin}}^{4}}x}{{{\operatorname{Cos}}^{2}}x}+\dfrac{{{\operatorname{Cos}}^{4}}x}{{{\operatorname{Sin}}^{2}}x}\] and then integrate them separately. Consider the first term, write ${{\operatorname{Sin}}^{2}}x$ as $1-{{\operatorname{Cos}}^{2}}x$ so that it becomes ${{\left( \operatorname{Sec}x-\operatorname{Cos}x \right)}^{2}}$ , expand this and you will get simplified terms which have direct integral formula . Similarly, we can evaluate the integral of the second term.