Question

Evaluate: $\int {{{\sin }^3}x{{\cos }^3}xdx} $.

Answer 1

Hint: In the above type of integration question first of all we will have to convert them by using trigonometric formulae in that form in which we can easily integrate them, so, we have to remember the sine and cosine sum angle formulae. After that use integration by substitution method to solve the problem.

Complete step by step solution:
In the above question, we have to find the integral of ${\sin ^3}x{\cos ^3}x$ which is in the multiplication form and we don’t know the integration of this kind. So, we will try to split it by the formula,
${\cos ^2}x = 1 - {\sin ^2}x$
So, by using the above formulae we can write the given expression is as follow;
$ \Rightarrow \int {{{\sin }^3}x{{\cos }^3}xdx} = \int {{{\sin }^3}x\left( {1 - {{\sin }^2}x} \right)\cos xdx} $
There are times when the given function is a little complicated and thus, making it difficult for us to integrate. To make it easy we use a different independent variable to make it easier to integrate. This is known as integration by substitution.
Now apply integration by substitution to integrate it.
Let us assume $\sin x = t$,
Differentiate it with respect to $x$,
$ \Rightarrow \cos x = \dfrac{{dt}}{{dx}}$
Cross multiply the terms,
$ \Rightarrow \cos xdx = dt$
Substitute the value in the equation,
$ \Rightarrow \int {{{\sin }^3}x{{\cos }^3}xdx} = \int {{t^3}\left( {1 - {t^2}} \right)dt} $
Open the brackets and multiply the terms,
$ \Rightarrow \int {{{\sin }^3}x{{\cos }^3}xdx} = \int {\left( {{t^3} - {t^5}} \right)dt} $
Integrate the terms,
$ \Rightarrow \int {{{\sin }^3}x{{\cos }^3}xdx} = \dfrac{{{t^4}}}{4} - \dfrac{{{t^6}}}{6} + c$
Now substitute back the value of $t$ in the equation,
$ \Rightarrow \int {{{\sin }^3}x{{\cos }^3}xdx} = \dfrac{{{{\sin }^4}x}}{4} - \dfrac{{{{\sin }^6}x}}{6} + c$

Hence, the integral of the given function in the above question will be $\dfrac{{{{\sin }^4}x}}{4} - \dfrac{{{{\sin }^6}x}}{6} + c$.

Note: Integration is a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points and many useful things. But it is easiest to start with finding the area under the curve of a function.
In such types of questions always choose substitution which makes integration simple, in above integration we choose $\sin x = t$, so it makes integration simple, then we easily integrate using some basic property of integration which is stated above, then simplify we will get the required answer.