Question

The value of integral $\int{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}$ is equal to:

Answer 1

Hint: As per the question we need to find the integration of some combination of trigonometric functions. Now, as it is an indefinite integral so we need to add some arbitrary constant to the integration obtained. We will make use of some identities in order to obtain the integration.

Complete step-by-step solution:
According to the question we need to find the value of the integration of $\int{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}$.
Now, after observing the given function that is to be integrated seems to be complicated and in order to simplify this, we will make use of some trigonometric identities.
Now, in order to integrate it, let $I=\int{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}$
Now, we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ , therefore using this we get
$\int{\dfrac{\left( {{\sin }^{4}}x-{{\cos }^{4}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}$
Now, again applying the same identity in the first braces of the numerator we get,
$\int{\dfrac{\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}$
Now, we know that
${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and also, we know that $\begin{align}
& {{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}^{2}}={{\sin }^{4}}x+{{\cos }^{4}}x+2{{\sin }^{2}}x{{\cos }^{2}}x \\
 & \Rightarrow {{\sin }^{4}}x+{{\cos }^{4}}x=1-2{{\sin }^{2}}x{{\cos }^{2}}x \\
\end{align}$
Using identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Therefore, the integration becomes
$\begin{align}
& \int{\dfrac{\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( 1-2{{\sin }^{2}}x{{\cos }^{2}}x \right)}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx} \\
 & \Rightarrow \int{\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)dx} \\
\end{align}$
Now, we also know that the cosine identity is $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$ .
So, now applying this in the integration by taking minus sign common we will get the following integral $-\int{\cos 2xdx}$ .This is the simplified integral obtained from the given integral and now, integrating this will give us the required result.
Therefore, integrating $-\int{\cos 2xdx}$ we get
 $\begin{align}
  & -\int{\cos 2xdx} \\
 & \Rightarrow -\dfrac{\sin 2x}{2}+C \\
\end{align}$ .
Therefore, the integration of $\int{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}$is $-\dfrac{\sin 2x}{2}+C$.

Note: We need to remember that while finding the integration of such large functions we need to first simplify them using various identities in order to integrate it easily otherwise we won’t be able to integrate and hence will give up there itself.