Question

How do you integrate \[{\sin ^3}2x\,dx\]?

Answer 1

Hint: Here in this question given an indefinite integral, we have to find the integrated value of given function. Since the given function is a trigonometric function by using the trigonometry ratios and trigonometry identities we simplify and then later integrate by using the standard formulas of integration. And by further simplification we get the required solution.

Complete step-by-step solution:
 We have to integrate the given function. The function is of the form of trigonometry we use trigonometry identities and we simplify the given function.
Now consider the given function
\[{\sin ^3}2x\,dx\]
Since the given trigonometric function is in the form of exponential, and this can be written as
\[ \Rightarrow {\sin ^2}2x.\sin 2x\,dx\]
As we know the trigonometry identity \[{\sin ^2}x\,dx = (1 - {\cos ^2}x)\], by applying this the above function is written as
\[ \Rightarrow (1 - {\cos ^2}2x)\sin 2x\,\]
On multiplying we get
\[ \Rightarrow \sin 2x - {\cos ^2}2x\sin 2x\,\]
We have to integrate this function.
Let we consider the first term as \[{I_1}\] and second term as \[{I_2}\]
So the function is written as \[I = {I_1} + {I_2}\]
On integrating \[{I_1}\],
\[ \Rightarrow {I_1} = \int {\sin 2x\,dx} \]
The integration formula of \[\sin x\] is \[ - \cos x\]
\[ \Rightarrow {I_1} = - \dfrac{1}{2}\cos 2x\]
On integrating \[{I_2}\],
\[ \Rightarrow {I_2} = - \int {{{\cos }^2}2x\sin 2x\,dx} \]
We integrate this function by substitution method. Because we have function and its derivative.
Let substitute \[u = \cos 2x\]then \[\dfrac{1}{2}\,du = \sin 2xdx\]. Therefore we have
\[ \Rightarrow {I_2} = - \dfrac{1}{2}\int {{u^2}\,du} \]
on integration above function
\[ \Rightarrow {I_2} = - \dfrac{{{u^3}}}{6} + C\]
Resubstitute the value of u, then we get
\[ \Rightarrow {I_2} = - \dfrac{{{{\cos }^3}2x}}{6} + C\]
Therefore the integral is written as
\[\int {{{\sin }^3}2x\,dx} = {I_1} + {I_2}\]
Substituting the values we get
\[\int {{{\sin }^3}2x\,dx} = - \dfrac{1}{2}\cos 2x - \dfrac{1}{6}{\cos ^3}2x + C\]
Hence we have integrated the given question and obtained the result.

Therefore the final answer \[\int {{{\sin }^3}2x\,dx} = - \dfrac{1}{2}\cos 2x - \dfrac{1}{6}{\cos ^3}2x + C\]

Note: While integrating the trigonometric functions, we simplify the trigonometric functions as much as possible by using the trigonometry ratios or by trigonometry identities. The integration by substitution is the easiest way to integrate. The function and its derivative must be present while substituting.