Question

The integral $\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{\left\{ {{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)-{{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right) \right\}dx}$ is equal to
A. 0
B. $\dfrac{\pi }{2}$
C. $\dfrac{3\pi }{2}$
D. $-\dfrac{\pi }{2}$

Answer 1

Hint: We first use the concept of inverse trigonometric functions and formulas like ${{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}x$ and ${{\sin }^{-1}}y+{{\cos }^{-1}}y=\dfrac{\pi }{2}$. We use these formulas to simplify the given equation $\left( {{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)-{{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right) \right)$. Then we use the definite integral formulas of $\int\limits_{-a}^{a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}$ for even function. We find the solution.

Complete step by step solution:
We have been given an equation for integral. We first simplify the equation.
We have ${{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)-{{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right)$.
We know that ${{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}x$.
We have that ${{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right)=\pi -{{\cos }^{-1}}\left( 3x-4{{x}^{3}} \right)$.
Replacing we get ${{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)-{{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right)={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)-\pi +{{\cos }^{-1}}\left( 3x-4{{x}^{3}} \right)$.
We also have the identity that ${{\sin }^{-1}}y+{{\cos }^{-1}}y=\dfrac{\pi }{2}$.
Therefore, ${{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)-\pi +{{\cos }^{-1}}\left( 3x-4{{x}^{3}} \right)=\dfrac{\pi }{2}-\pi =-\dfrac{\pi }{2}$.
Now we complete the integral
$\begin{align}
  & \int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{\left\{ {{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)-{{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right) \right\}dx} \\
 & =\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{\left\{ -\dfrac{\pi }{2} \right\}dx} \\
\end{align}$
Now we use the concept of definite integral where $\int\limits_{-a}^{a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}$ if the function is even function and $\int\limits_{-a}^{a}{f\left( x \right)dx}=0$ if the function is odd function.
For our integral the function is even. So, $-\dfrac{\pi }{2}\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{dx}=-\pi \int\limits_{0}^{\dfrac{1}{2}}{dx}=-\pi \left[ x \right]_{0}^{\dfrac{1}{2}}=\dfrac{-\pi }{2}$.
The integral $\int\limits_{-\dfrac{1}{2}}^{\dfrac{1}{2}}{\left\{ {{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right)-{{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right) \right\}dx}$ is $\dfrac{-\pi }{2}$.
The correct option is D.

Note: We need to remember that we can also change the variable of the function where we take $x=\sin \alpha $ and $x=\cos \beta $ for different equations. We simplify the equation and find differential forms to find the solution.