Question

How do you find the integral of \[{{\cos }^{-1}}x\left( x \right)\]?

Answer 1

Hint: From the question we have been asked to find the integral of a given inverse trigonometric function. For solving this question we will use the concept of integration and its applications. We use the formulae of integration by parts which is \[\Rightarrow \int{udv=uv-\int{vdu}}\] and we will simplify further using basic mathematical operations like multiplications, addition to solve this question.

Complete step by step answer:
Here let us assume the given expression as follows,
\[I=\int{{{\cos }^{-1}}x\left( x \right)dx}\]
Now we will use the integration by parts formula, which is \[\Rightarrow \int{udv=uv-\int{vdu}}\] to the above equation. so, we get the equation reduced as follows.
\[\Rightarrow I={{\cos }^{-1}}x\int{xdx-\int{\left( \dfrac{d}{dx}\left( {{\cos }^{-1}}x \right)\int{xdx} \right)dx}}\]
Here we use the formulae \[\Rightarrow \int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\] and \[\Rightarrow \int{{{\cos }^{-1}}x}=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}\]
\[\Rightarrow I={{\cos }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}\times \dfrac{{{x}^{2}}}{2}dx}\]
\[\Rightarrow I={{\cos }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx}\]
Now we will assume the above equation as follows.
\[\Rightarrow I={{\cos }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+\dfrac{1}{2}{{I}_{1}}\]
Where \[{{I}_{1}}=\int{\dfrac{{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx}\]
Now we use the substitution method and substitute \[x=\sin u\]. So, we get the \[dx=\cos udu\] as derivative of \[\dfrac{d}{dx}\sin x=\cos x\].
So, we will take the above equation and substitute the new x values and do the calculation.
So, we get as follows.
\[\Rightarrow {{I}_{1}}=\int{\dfrac{{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx}\]
\[\Rightarrow {{I}_{1}}=\int{\dfrac{{{\sin }^{2}}u}{\sqrt{1-{{\sin }^{2}}u}}\cos udu}\]
Here we use the formulae \[\Rightarrow 1-{{\sin }^{2}}x={{\cos }^{2}}x\].
\[\Rightarrow {{I}_{1}}=\int{\dfrac{{{\sin }^{2}}u}{\sqrt{{{\cos }^{2}}u}}\cos udu}\]
\[\Rightarrow {{I}_{1}}=\int{\dfrac{{{\sin }^{2}}u}{\cos u}\cos udu}\]
\[\Rightarrow {{I}_{1}}=\int{{{\sin }^{2}}udu}\]
Here we use the formulae \[\Rightarrow \cos 2x=1-2{{\sin }^{2}}x\].
\[\Rightarrow {{I}_{1}}=\int{\dfrac{1-\cos 2u}{2}du}\]
\[\Rightarrow {{I}_{1}}=\dfrac{1}{2}\left( u-\dfrac{\sin 2u}{2} \right)+c\]
\[\Rightarrow {{I}_{1}}=\dfrac{1}{2}\left( u-\sin u\cos u \right)+c\]
\[\Rightarrow {{I}_{1}}=\dfrac{1}{2}\left( u-\sin u\sqrt{1-{{\sin }^{2}}u} \right)+c\]
Now we will substitute back the values of x we took previously. So, we get the equation reduced as follows.
\[\Rightarrow {{I}_{1}}=\dfrac{1}{2}\left( {{\sin }^{-1}}x-x\sqrt{1-x} \right)+c\]
Now we will take this and substitute in the equation \[\Rightarrow I={{\cos }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+\dfrac{1}{2}{{I}_{1}}\]. So, we get the equation reduced as follows.
\[\Rightarrow I={{\cos }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+\dfrac{1}{2}{{I}_{1}}\]
\[\Rightarrow I={{\cos }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+\dfrac{1}{2}\left( \dfrac{1}{2}\left( {{\sin }^{-1}}x-x\sqrt{1-x} \right)+c \right)\]
\[\Rightarrow I={{\cos }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)+\left( \dfrac{1}{4}\left( {{\sin }^{-1}}x-x\sqrt{1-x} \right) \right)+C\]

Note: Students must have good knowledge in the concept of inverse trigonometry and trigonometry and its applications. We must also know the formulae in integrations to solve this question. The formulae we required to solve this question are as follows.
\[\Rightarrow \int{udv=uv-\int{vdu}}\]
\[\Rightarrow \int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\] and \[\Rightarrow \int{{{\cos }^{-1}}x}=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}\]
\[\Rightarrow 1-{{\sin }^{2}}x={{\cos }^{2}}x\]
\[\Rightarrow \cos 2x=1-2{{\sin }^{2}}x\]