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Question: Select the correct option: \(\int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {{x^3}{{\sin }^4}xdx} \) is...

Select the correct option: ∫−π4π4x3sin⁡4xdx\int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {{x^3}{{\sin }^4}xdx} is equal to:
A. 1
B. 2
C. 0
D. None of these

Explanation

Solution

We will first prove that the function inside the integral is an odd function. After that, we will use an identity for definite integral of odd functions, we will get the required answer.

Step-By-Step answer:
Let us say that f(x)=x3sin⁡4xf(x) = {x^3}{\sin ^4}x
If we replace x by –x in both sides, we will then obtain:-
⇒f(−x)=(−x)3sin⁡4(−x)\Rightarrow f( - x) = {( - x)^3}{\sin ^4}( - x)
Now, we know that the sine function is negative in the fourth quadrant and thus we also have: sin⁡(−x)=−sin⁡x\sin ( - x) = - \sin x.
Using this, in the above expression of f(-x), we will then obtain:-
⇒f(−x)=−x3(−sin⁡x)4\Rightarrow f( - x) = - {x^3}{( - \sin x)^4}
Simplifying it, we will then obtain:-
⇒f(−x)=−x3sin⁡4x\Rightarrow f( - x) = - {x^3}{\sin ^4}x
We can clearly observe that f(x)=x3sin⁡4x=−f(−x)f(x) = {x^3}{\sin ^4}x = - f( - x)
Hence, we can conclude that f(x) is an odd function.
Now, we also know that if g(x) is in odd function, then:
∫−aag(x)dx=0\int_{ - a}^a {g(x)dx = 0}
Hence, using this in f(x)=x3sin⁡4xf(x) = {x^3}{\sin ^4}x, we will then obtain:-
⇒∫−π4π4x3sin⁡4xdx=0\Rightarrow \int_{ - \dfrac{\pi }{4}}^{\dfrac{\pi }{4}} {{x^3}{{\sin }^4}xdx} = 0

Hence, the correct option is (C) 0.

Note: The students must note that, they need not cram the fact that if we have g(x) as an odd function, then∫−aag(x)dx=0\int_{ - a}^a {g(x)dx = 0} .
But they may also break it like this:-
⇒∫−aag(x)dx=∫−a0g(x)dx+∫0ag(x)dx\Rightarrow \int_{ - a}^a {g(x)dx = } \int_{ - a}^0 {g(x)dx + \int_0^a {g(x)dx} }
We can rewrite the above expression as follows:-
⇒∫−aag(x)dx=∫0−a−g(x)dx+∫0ag(x)dx\Rightarrow \int_{ - a}^a {g(x)dx = } \int_0^{ - a} { - g(x)dx + \int_0^a {g(x)dx} }
Now, since we know that g(x) is an odd function, so we have: g(x) = - g(-x)
Putting this in the above expression, we will get:-
⇒∫−aag(x)dx=∫0−ag(−x)dx+∫0ag(x)dx\Rightarrow \int_{ - a}^a {g(x)dx = } \int_0^{ - a} {g( - x)dx + \int_0^a {g(x)dx} }
We can also rewrite this as:-
⇒∫−aag(x)dx=−∫0ag(x)dx+∫0ag(x)dx\Rightarrow \int_{ - a}^a {g(x)dx = } - \int_0^a {g(x)dx + \int_0^a {g(x)dx} }
Clubbing the like integral on the right hand side of the above expression, we will thus obtain:-
⇒∫−aag(x)dx=0\Rightarrow \int_{ - a}^a {g(x)dx = } 0
Hence, we have the proof for the identity / theorem we used in the above result.
Let us learn some facts about odd functions:-
Geometrically, the graph of an odd function has a rotational symmetry with respect to the origin, which means that its graph remains intact after the rotation of 180 degrees about the origin.
If some function is both an even and odd function, then it is zero everywhere it is defined.