Integrate the following trigonometric function:
\[\int {\dfrac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx\]
A) \[\dfrac{1}{2}\sin 2x + C\]
B) \[ - \dfrac{1}{2}\sin 2x + C\]
C) \[ - \dfrac{1}{2}\sin x + C\]
D) \[ - {\sin ^2}x + C\]

Hint: Here first we will expand the numerator using various trigonometric identities and then after cancelling the terms we will integrate the value inside the integral.
The identities used are:-
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[{\cos ^2}x + {\sin ^2}x = 1\]
\[{\cos ^2}x - {\sin ^2}x = \cos 2x\]

Complete step-by-step answer:
Let
\[I = \int {\dfrac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx\]………………………..(1)
Now considering the numerator we get:-
\[ = {\sin ^8}x - {\cos ^8}x\]
It can also be written as:-
\[ = {\left( {{{\sin }^4}x} \right)^2} - {\left( {{{\cos }^4}x} \right)^2}\]
Now we will apply the following identity:-
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
On applying the identity in the above expression we get:-
\[ = \left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^4}x - {{\cos }^4}x} \right)\]
Now again we can write \[{\sin ^4}x\] as \[{\left( {{{\sin }^2}x} \right)^2}\] and \[{\cos ^4}x\] as \[{\left( {{{\cos }^2}x} \right)^2}\]
Hence substituting the values we get:-
\[ = \left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left[ {{{\left( {{{\sin }^2}x} \right)}^2} - {{\left( {{{\cos }^2}x} \right)}^2}} \right]\]
Now again applying the following identity:-
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
We get:-
\[ = \left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left[ {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right)} \right]\]
Now as we know that:
\[{\cos ^2}x + {\sin ^2}x = 1\]
Hence substituting the value we get:-
\[
   = \left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left[ {\left( 1 \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right)} \right] \\
   \Rightarrow \left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right) \\
 \]
Now we know that:-
\[{\cos ^2}x - {\sin ^2}x = \cos 2x\]
Hence, \[{\sin ^2}x - {\cos ^2}x = - \cos 2x\]
Hence substituting this value in above expression we get:-
\[ = \left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( { - \cos 2x} \right)..........................\left( 2 \right)\]
Now since we know that:-
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Hence, \[{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} = {\sin ^4}x + {\cos ^4}x + 2{\sin ^2}x{\cos ^2}x\]
Now evaluating the value of \[{\sin ^4}x + {\cos ^4}x\] from the above equation we get:-
\[{\sin ^4}x + {\cos ^4}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\]
Now we already know that:-
\[{\cos ^2}x + {\sin ^2}x = 1\]
Hence substituting the value we get:-
\[
  {\sin ^4}x + {\cos ^4}x = {\left( 1 \right)^2} - 2{\sin ^2}x{\cos ^2}x \\
   \Rightarrow {\sin ^4}x + {\cos ^4}x = 1 - 2{\sin ^2}x{\cos ^2}x.....................\left( 3 \right) \\
 \]
Now substituting the value of equation 3 in equation 2 we get:-
\[ = \left( {1 - 2{{\sin }^2}x{{\cos }^2}x} \right)\left( { - \cos 2x} \right)\]
Now substituting this value in equation 1 we get:-
\[I = \int {\dfrac{{\left( {1 - 2{{\sin }^2}x{{\cos }^2}x} \right)\left( { - \cos 2x} \right)}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx\]
Now cancelling the required terms we get:-
\[
  I = \int { - \cos 2xdx} \\
   \Rightarrow I = - \int {\cos 2xdx} \\
 \]
Now we know that:
\[\int {\cos xdx = \sin x + C} \]
Hence, \[\int {\cos 2xdx} = \dfrac{1}{2}\sin 2x + C\]
Substituting the value in above expression we get:-
\[I = - \dfrac{1}{2}\sin 2x + C\]
Therefore, the value of \[\int {\dfrac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx\] is \[ - \dfrac{1}{2}\sin 2x + C\]

So, the correct answer is “Option B”.

Note: Students should note that in such types of questions they need to cancel maximum terms possible in order to make the question easy to integrate and use the known identities to simplify it.
Also they should note that:
\[\int {\cos xdx = \sin x + C} \] as integration is defined as the anti-derivative of a function.
Students may make mistakes while using the identities so, identities applied should be correct.