Solve the integration $\int{\left( {{\sin }^{4}}x-{{\cos }^{4}}x \right)dx}$ is equal to:
(A). $-\dfrac{\cos 2x}{2}+c$
(B). $-\dfrac{\sin 2x}{2}+c$
(C). $\dfrac{\sin 2x}{2}+c$
(D). $\dfrac{\cos 2x}{2}+c$

Hint: First take the difference of 4th power terms as the difference of 2 squares. By general identity in algebra convert into the product of 2 degrees 2 equations.
Now you can use trigonometry questions to simplify it further. After simplifying you’ll end up with integration of the 2nd degree trigonometric equation. Now use the trigonometry formula to convert it into a 1st degree equation. Now you can directly integrate like a normal integration use the formulae:
$\begin{align}
  & {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) \\
 & {{\cos }^{2}}x+{{\sin }^{2}}x=1;{{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x \\
\end{align}$

Complete step-by-step solution -
Given integration in the question is written as follows:
$\int{\left( {{\sin }^{4}}x-{{\cos }^{4}}x \right)dx}$
Now, let us assume this integration to be as I.
$I=\int{\left( {{\sin }^{4}}x-{{\cos }^{4}}x \right)dx}$
By normal algebra we can write those terms in the form:
${{\sin }^{4}}x={{\left( {{\sin }^{2}}x \right)}^{2}};{{\cos }^{4}}x={{\left( {{\cos }^{2}}x \right)}^{2}}$
By substituting these equations into the ‘I’, we get it as:
$I=\int{{{\left( {{\sin }^{2}}x \right)}^{2}}-{{\left( {{\cos }^{2}}x \right)}^{2}}dx}$
By normal algebra, we get the relation in terms of a,b:
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
By substituting this relation to our integration, we get it as –
$I=\int{\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)dx}$
By basic trigonometry, we have an identity defined as follows:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
By substituting this into our integration, we get it as:
$I=\int{\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)dx}$
By basic trigonometry, we have an identity defined as follows:
${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$ .
By substituting this into our integration, we get it as:
$I=\int{\cos 2x.dx}$
By assuming $2x=t$ we get $dx=\dfrac{dt}{2}$ by substituting, we get:
$I=-\dfrac{1}{2}\int{\cos t.dt}$
By basic integration, we get it as: $I=-\dfrac{1}{2}\sin t+C$
By substituting the value of t back into I. we get:
$I=-\dfrac{\sin 2x}{2}+C$
Therefore, option (b) is correct.

Note: Be careful while using ${{a}^{2}}-{{b}^{2}}$ as you must write $a-b$ if you write it reverse you will get $\dfrac{\sin 2x}{2}$ as answer which is also present in options. After getting ${{\sin }^{2}}x-{{\cos }^{2}}x$ don’t substitute $\cos 2x$ directly. Look carefully it is $-\cos 2x$. Students confuse this step. The idea of decreasing the degree of the equation is very important in the integration as it makes the solution simple.