Question

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

Answer 1

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
$\Rightarrow 2I=\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}-\int\limits_{-1}^{1}{\left( {{\sin }^{7}}x \right)dx}$

& \Rightarrow 2I=0 \\\ & \therefore I=0 \\\ \end{aligned}$$ Thus, we get $$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$$.