Question

How do you find the antiderivative of \[\int{\left( {{\sin }^{7}}x \right)dx}\] from \[[-1,1]\]?

Answer 1

Hint: To solve the given integral, we need to know some of the properties of integration. The property we will be using to solve this integration is \[\int\limits_{a}^{b}{f(x)}=\int\limits_{a}^{b}{f(a+b-x)}\]. This property is used as follows,
Let’s say \[I=\int\limits_{a}^{b}{f(x)}\], using the above property we get \[I=\int\limits_{a}^{b}{f(a+b-x)}\]. Adding both of the above integrals, we get \[2I=\int\limits_{a}^{b}{f(x)}+\int\limits_{a}^{b}{f(a+b-x)}\].

Complete step-by-step answer:
We are asked to find the antiderivative of \[\int{\left( {{\sin }^{7}}x \right)dx}\] from \[[-1,1]\]. This means we have to find the value of the definite integral \[\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}\]. We can use the integration property \[\int\limits_{a}^{b}{f(x)}=\int\limits_{a}^{b}{f(a+b-x)}\]. Using this property, we get \[\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}\left( 1-1-x \right) \right)dx}\].
Let’s say \[I=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}\]. From above, we can also say that \[I=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}\left( 1-1-x \right) \right)dx}\]. Adding both integrals, we get
\[2I=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}+\int\limits_{-1}^{1}{\left( {{\sin }^{7}}\left( 1-1-x \right) \right)dx}\]
Simplifying the above equation, we get
\[\Rightarrow 2I=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}+\int\limits_{-1}^{1}{\left( {{\sin }^{7}}\left( -x \right) \right)dx}\]
We know that, \[\sin (-x)=-\sin x\]. Using the property in the above equation, we get
\[\Rightarrow 2I=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}+\int\limits_{-1}^{1}{\left( {{\left( -\sin x \right)}^{7}} \right)dx}\]
Simplifying the above equation, we get
\[\Rightarrow 2I=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}+\int\limits_{-1}^{1}{-\left( {{\sin }^{7}}x \right)dx}\]
As we know that constants can be taken out of the integration, taking \[-1\] out of the integration from the second integral in the right-hand side of the above equation, we get
\[\Rightarrow 2I=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}-\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}\]
\[\begin{align}
  & \Rightarrow 2I=0 \\
 & \therefore I=0 \\
\end{align}\]
Thus, we get \[I=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}=0\].

Note: We can make a general property from the above question, to solve these types of problems as follows:
Let’s say \[f(x)\] is an odd function, from this, we can say that \[f(-x)=f(x)\]. Then the value of the definite integral \[\int\limits_{-a}^{a}{f(x)dx}\] equals 0.
For the given question, we have \[f(x)={{\sin }^{7}}x\] which is an odd function. Hence, using the above property, we can say that \[\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}=0\].