Question

What is the value of the integral \[\int\limits_{ - 1}^1 {{{\sin }^{11}}x} dx\]?
A. \[\dfrac{{10}}{{11}} \cdot \dfrac{8}{9} \cdot \dfrac{6}{7} \cdot \dfrac{4}{5} \cdot \dfrac{2}{3}\]
B. \[\dfrac{{10}}{{11}} \cdot \dfrac{8}{9} \cdot \dfrac{6}{7} \cdot \dfrac{4}{5} \cdot \dfrac{2}{3} \cdot \dfrac{\pi }{2}\]
C. 1
D. 0

Answer 1

Hint: Here, a definite integral is given. First, check whether the function present in the integral is an even or an odd function. If the function is odd, then the value of that integral is 0. If the function is even, then solve the integral by applying the integration rule \[\int\limits_{ - a}^a {f\left( x \right)} dx = 2\int\limits_0^a {f\left( x \right)} dx\]. In the end, simplify the integral to get the required answer.



Formula Used:\[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function. Means, \[f\left( { - x} \right) = - f\left( x \right)\]
\[\int\limits_{ - a}^a {f\left( x \right) dx} = 2\int\limits_0^a {f\left( x \right) dx} \], if the function \[f\left( x \right)\] is an even function. Means, \[f\left( { - x} \right) = f\left( x \right)\]



Complete step by step solution:The given definite integral is \[\int\limits_{ - 1}^1 {{{\sin }^{11}}x} dx\].

Let consider,
\[f\left( x \right) = {\sin ^{11}}x\]
Now we have to check whether the above function is an odd function or an even function.
So, let’s calculate the value of \[f\left( { - x} \right)\].
\[f\left( { - x} \right) = {\sin ^{11}}\left( { - x} \right)\]
Apply the trigonometric identity \[{\sin ^n}\left( { - x} \right) = {\left( { - \sin x} \right)^n}\].
\[f\left( { - x} \right) = {\left( { - \sin x} \right)^{11}}\]
Since, \[11\] is an odd number.
So, we get
\[f\left( { - x} \right) = - {\sin ^{11}}x\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
Therefore, the function \[f\left( x \right) = {\sin ^{11}}x\] is an odd function.
Now apply the property of the definite integral \[\int\limits_{ - a}^a {f\left( x \right) dx} = 0\], if the function \[f\left( x \right)\] is an odd function.
We get,
\[\int\limits_{ - 1}^1 {{{\sin }^{11}}x} dx = 0\]



Option ‘D’ is correct



Note: Sometimes students get confused and write \[{\sin ^n}\left( { - x} \right) = {\left( {\sin x} \right)^n}\]. Which is a wrong formula. Because of that, \[{\sin ^{11}}x\] will be considered as an even function and they get a different solution. The correct formula is \[{\sin ^n}\left( { - x} \right) = {\left( { - \sin x} \right)^n}\].