Integrate the function \[x{\cos ^{ - 1}}x\]

Hint: The simple meaning of trigonometry is calculations of triangles.
As we know that-
 \[\cos \theta = \dfrac{{Base}}{{Hypotenuse}}\]
Differentiation of \[\cos \theta = - \sin \theta \]
Also, \[\sin \theta = \sqrt {1 - {{\cos }^2}\theta } \]
 \[{\cos ^2}\theta = 1 + \cos 2\theta \]
 \[\sin 2\theta = 2\cos \theta \sin \theta \]
By using these identities we will solve the sum.
In this question we use integration by parts, by taking the equation and considering one as u and another as v, we can solve this question and will get the answer.

Complete step-by-step answer:
For solving this question, we have to take the integration by parts.
 So, first consider, \[I = \int {x{{\cos }^{ - 1}}} x\] \[dx\]
Also, \[I = \int {{{\cos }^{ - 1}}x.x} \] \[dx\]
Integrating by parts
  \[I\] = \[{\cos ^{ - 1}}x.\dfrac{{{x^2}}}{2} - \int {\dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}} \dfrac{{{x^2}}}{2}dx\]
 \[I = \dfrac{{{x^2}}}{2}{\cos ^{ - 1}}x + \dfrac{1}{2}\] \[{I_1}\] ………….(A)
 \[I\] = \[\dfrac{{{x^2}}}{2}{\cos ^{ - 1}}x + \dfrac{1}{2}\int {\dfrac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} dx\]
Let \[{I_1}\] = \[\int {\dfrac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} dx\]
Put \[x = \cos \theta \] ,
If we take differentiation of the above term then we get-
  \[dx = - \sin \theta d\theta \]
So, \[{I_1}\] = \[\int {\dfrac{{{{\cos }^2}\theta }}{{\sqrt {1 - {{\cos }^2}\theta } }}} \left( { - \sin \theta } \right)d\theta \]
 \[{I_1}\] = - \[\int {{{\cos }^2}\theta } d\theta \]
 \[{I_1}\] = \[\int {\left( {\dfrac{{1 + \cos 2\theta }}{2}} \right)} d\theta \]
 \[{I_1}\] = \[ - \dfrac{1}{2}\theta - \dfrac{1}{2}.\dfrac{{\sin 2\theta }}{2}\] ∴ \[\sin 2\theta = 2\cos \theta \sin \theta \]
As we know the identity, so by using this in the above equation, we get-
  \[{I_1}\] = \[ - \dfrac{1}{2}\theta - \dfrac{1}{2}.\dfrac{{2\sin \theta \cos \theta }}{2} + {c_1}\]
  \[{I_1}\] = \[ - \dfrac{1}{2}\theta - \dfrac{1}{2}\sin \theta \cos \theta + {c_1}\]
  \[\cos \theta = x\] , \[\sin \theta = \sqrt {1 - {{\cos }^2}\theta } \]
  So, \[\sin \theta = \sqrt {1 - {x^2}} \]
  \[{I_1}\] = \[ - \dfrac{1}{2}{\cos ^{ - 1}}x - \dfrac{1}{2}x\sqrt {1 - {x^2}} + {c_1}\]
Substitute the value of \[{I_1}\] in equation (A), after substituting all the values in equation (A) we will get-
So, \[I\] = \[\dfrac{{{x^2}}}{2}{\cos ^{ - 1}}x - \dfrac{1}{4}{\cos ^{ - 1}}x - \dfrac{1}{4}x\sqrt {1 - {x^2}} + c\]
Hence, this will be the resulting answer.
So, the correct answer is “ \[\dfrac{{{x^2}}}{2}{\cos ^{ - 1}}x - \dfrac{1}{4}{\cos ^{ - 1}}x - \dfrac{1}{4}x\sqrt {1 - {x^2}} + c\] ”.

Note: In this question, we are using identities, by using that question will be easy to solve. By choosing the correct identity for solving the question remember to check the a and x value. Also remember to check the positive and negative signs. The simple meaning of trigonometry is calculations of triangles. Also, in physics, trigonometry is used to find the components of vectors and also in projectile motion have a lot of application of trigonometry.