Question

How do you find the integral of \[\cos^{4}\left( 2x \right).\sin^{3}\left( 2x \right)\] ?

Answer 1

Hint:In this question, we need to find the integral of \[\cos^{4}\left( 2x \right).\sin^{3}\left( 2x \right)\]. Integration is nothing but the opposite of derivative. Integration of derivativation of a function is equal to its original function. The inverse of differentiation is also known as integral. The symbol `\[\int\]’ is known as the sign of integration and is used in the process of integration . Let us consider the given expression as \[I\] . For this we will use the trigonometric identity \[\sin^{2}\left( x \right) = \left( 1 - \cos^{2}\left( x \right) \right)\] . After integrating, then we need to try to do some rearrangements of terms and hence we can find the integral of the given expression.

Complete step by step answer:
Given, \[\cos^{4}\left( 2x \right).\sin^{3}\left( 2x \right)\ dx\]
Here we need to find the integral of \[\cos^{4}\left( 2x \right).\sin^{3}\left( 2x \right)\ dx\]
Let us consider the given expression as \[I\] .
\[\Rightarrow \ I = \int \cos^{4}\left( 2x \right).\sin^{3}\left( 2x \right)\ dx\]
We have rewrite \[\sin^{3}\left( 2x \right)\] as \[\sin(2x) \times \ \sin^{2}\left( 2x \right)\]
By rewriting,
We get,
\[\Rightarrow \ I = \int \cos^{4}\left( 2x \right).\sin\left( 2x \right) \times \ \sin^{2}\left( 2x \right)\]
We know that
\[\sin^{2}\left( x \right) = \left( 1 - \cos^{2}\left( x \right) \right)\] ,

By using this identity ,
We get,
\[\Rightarrow \ I = \int \cos^{4}\left( 2x \right).\sin\left( 2x \right) \times \left( 1 - \cos^{2}\left( x \right) \right)\]
On multiplying the terms,
We get,
\[\Rightarrow \ I = \int \cos^{4}\left( 2x \right).\sin\left( 2x \right) - \cos^{4}\left( 2x \right).\cos^{2}\left( 2x \right).\sin\left( 2x \right)\]
On simplifying,
We get,
\[I = \int \cos^{4}\left( 2x \right).\sin\left( 2x \right) - \cos^{6}\left( 2x \right).\ \sin\left( 2x \right)\] ••• (1)

We know that,
\[\dfrac{d}{dx}\left( \alpha \cos^{n}\left( 2x \right) \right) = \ - 2n\alpha \cos^{(n – 1)}\ \left( 2x \right)\sin\left( 2x \right)\]
On dividing both sides by \[\frac{1}{2n\alpha}\] ,
We get,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{2n}\left( \cos^{n}\left( 2x \right) \right) \right) = - \cos^{(n – 1)}\left( 2x \right)\sin\left( 2x \right)\]
We know that the inverse of differentiation is known as integral.
On taking inverse on both sides,
We get,
\[\int \cos^{\left( n – 1 \right)}\left( 2x \right)\sin\left( 2x \right) = - \dfrac{1}{2n}\left( \cos^{n}\left( 2x \right) \right) + c\]
Where \[c\] is the constant of integration.

Thus on using this expression our equation (1) becomes,
Equation (1) is \[\int \cos^{4}\left( 2x \right).\sin\left( 2x \right) - \cos^{6}\left( 2x \right).\ \sin\left( 2x \right)\]
Now on applying \[\int \cos^{(n – 1)}\left( 2x \right)\sin\left( 2x \right) = - \dfrac{1}{2n}\left( \cos^{n}\left( 2x \right) \right) + c\] ,
We get,
\[I \Rightarrow \ \int \cos^{(5 – 1)}\left( 2x \right).\sin\left( 2x \right) - \cos^{(7 – 1)}\left( 2x \right).\ \sin\left( 2x \right) = - \dfrac{1}{2\left( 5 \right)}\left( \cos^{5}\left( 2x \right) \right) - \left( - \dfrac{1}{2\left( 7 \right)}\left( \cos^{7}\left( 2x \right) \right) \right) + c\]
Now on simplifying,
We get,
\[\therefore I = - \dfrac{1}{10}\left( \cos^{5}\left( 2x \right) \right) + \dfrac{1}{14}\left( \cos^{7}\left( 2x \right) \right) + c\]

Thus we get the integral of \[\cos^{4}\left( 2x \right).\ \sin^{3}\left( 2x \right)\ dx\] is \[- \dfrac{1}{10}\left( \cos^{5}\left( 2x \right) \right) + \dfrac{1}{14}\left( \cos^{7}\left( 2x \right) \right) + c\].

Note:We must have a strong grip over the integral calculus to solve such a complex question of indefinite integration. The given expression can be simplified by making use of trigonometric identities. We should be careful in choosing the correct substitution, that is also very important. We should not get confused with the integration and differentiation of cosine and sine functions. Also, we must be careful while doing the calculations in order to get the final answer.