Question

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

Answer 1

We have to evaluate the value of $\int\limits_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {{{\sin }^{ - 1}}\left( {3x - 4{x^3}} \right) - {{\cos }^{ - 1}}\left( {4{x^3} - 3x} \right)dx}$. We will write a simplified expression for the given terms using the properties of trigonometry. Then, we will do integration of the expression by applying the formulas of integration. After integration, we will put the limits and then solve it to get the required answer.

Complete step by step solution:
It is given that we have to evaluate the following integral, $\int\limits_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {{{\sin }^{ - 1}}\left( {3x - 4{x^3}} \right) - {{\cos }^{ - 1}}\left( {4{x^3} - 3x} \right)dx}$We will simplify the terms separately.
Put $x = \sin y$ in the term, ${\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$, then $y = {\sin ^{ - 1}}x$
$\Rightarrow {\sin ^{ - 1}}\left( {3\sin y - 4{{\sin }^3}y} \right) \\\ \Rightarrow {\sin ^{ - 1}}\left( {\sin 3y} \right) \\\ \Rightarrow 3y \\\$
On substituting the value of $y$, we will get,
$= 3{\sin ^{ - 1}}x$
Similarly, let $x = \cos z$ in the term ${\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)$, then $z = {\cos ^{ - 1}}x$
$\Rightarrow {\cos ^{ - 1}}\left( {4{{\cos }^3}z - 3\cos z} \right) \\\ \Rightarrow {\cos ^{ - 1}}\left( {\cos 3z} \right) \\\ \Rightarrow 3z \\\$
On substituting the value of $z$, we will get,
$= 3{\cos ^{ - 1}}x$
Now, the given expression can be written as,
$\Rightarrow \int\limits_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\left( {3{{\sin }^{ - 1}}x - 3{{\cos }^{ - 1}}x} \right)dx} \\\ \Rightarrow 3\int\limits_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} {\left( {{{\sin }^{ - 1}}x - {{\cos }^{ - 1}}x} \right)dx} \\\$
We know that $\int {{{\sin }^{ - 1}}xdx = x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} }$ and $\int {{{\cos }^{ - 1}}xdx = x{{\cos }^{ - 1}}x - \sqrt {1 - {x^2}} }$
When we integrate the above expression, we will get,
$= 3\left[ {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} - x{{\cos }^{ - 1}}x + \sqrt {1 - {x^2}} } \right]_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} \\\ = 3\left[ {x{{\sin }^{ - 1}}x - x{{\cos }^{ - 1}}x + 2\sqrt {1 - {x^2}} } \right]_{ - \dfrac{1}{2}}^{\dfrac{1}{2}} \\\$
When we will substitute the limits, we will get,
$= 3\left[ {\dfrac{1}{2}{{\sin }^{ - 1}}\left( {\dfrac{1}{2}} \right) - \dfrac{1}{2}{{\cos }^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2\sqrt {1 - \left( {\dfrac{1}{4}} \right)} } \right] - 3\left[ {\left( { - \dfrac{1}{2}} \right){{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right) - \left( { - \dfrac{1}{2}} \right){{\cos }^{ - 1}}\left( { - \dfrac{1}{2}} \right) + 2\sqrt {1 - \dfrac{1}{4}} } \right] \\\ = 3\left[ {\dfrac{1}{2}\left( {\dfrac{\pi }{6}} \right) - \dfrac{1}{2}\left( {\dfrac{\pi }{3}} \right) + \sqrt 3 } \right] - 3\left[ { - \dfrac{1}{2}\left( { - \dfrac{\pi }{6}} \right) + \dfrac{1}{2}\left( {\pi - \dfrac{\pi }{3}} \right) + \sqrt 3 } \right] \\\ = 3\left[ {\dfrac{\pi }{{12}} - \dfrac{\pi }{6} + \sqrt 3 } \right] - 3\left[ {\dfrac{\pi }{{12}} + \dfrac{\pi }{3} + \sqrt 3 } \right] \\\ = 3\left[ {\dfrac{\pi }{{12}} - \dfrac{\pi }{6} + \sqrt 3 - \dfrac{\pi }{{12}} - \dfrac{\pi }{3} - \sqrt 3 } \right] \\\ = 3\left[ { - \dfrac{\pi }{6} - \dfrac{\pi }{3}} \right] \\\ = - \dfrac{{3\pi }}{2} \\\$

Hence, option C is correct.

Note:
We have substituted $x = \sin y$in first term as we get an expression $3\sin y - 4{\sin ^3}y$ which is equal to $\sin 3y$ and $x = \cos z$ in second term as $4{\cos ^3}z - 3\cos z$ is equal to $\cos 3z$. While substituting the limits, we will first put the upper limit for the first term and then the lower limit.