Question

Integrate the function: \(xcos^{-1} x\)

Answer 1

The correct answer is: \(=\frac{(2x^2-1)}{4}cos^{-1}x-\frac{x}{4}\sqrt{1-x^2}+C\)
Let \(I=∫xcos^{-1} x. dx\) Taking \(cos^{−1} x\) as first function and \(x\) as second function and integrating by parts,we
obtain
\(I=cos^{-1}x∫x.dx-∫[{(\frac{d}{dx}cos^{-1}x)∫xdx}]dx\)
\(=cos^{-1}x\, \frac{x^2}{2}-∫\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2} dx\)
\(=\frac{x^2cos^{-1}}{2}-\frac{1}{2}∫\frac{1-x^2-1}{\sqrt{1-x^2}} dx\)
\(=\frac{x^2cos^{-1}}{2}-\frac{1}{2}\int{\sqrt{1-x^2}+(\frac{-1}{\sqrt{1-x^2}})}dx\)
\(=\frac{x^2cos^{-1}}{2}-\frac{1}{2}\int{\sqrt{1-x^2}+\frac{1}{2}\int(\frac{-1}{\sqrt{1-x^2}})}dx\)
\(=\frac{x^2cos^{-1}}{2}-\frac{1}{2}I_1-\frac{1}{2}cos^{-1}x...(1)\)
Where,\(I_1=∫\sqrt{1-x^2} dx\)
\(⇒I_1=x\sqrt{1-x^2}-∫\frac{d}{dx}\sqrt{1-x^2}∫x.dx\)
\(⇒I_1=x\sqrt{1-x^2}-∫\frac{-2x}{2\sqrt{1-x^2}}.x dx\)
\(⇒I_1=x\sqrt{1-x^2}-∫\frac{-x}{\sqrt{1-x^2}}.x dx\)
\(⇒I_1=x\sqrt{1-x^2}-∫\frac{1-x^2-1}{\sqrt{1-x^2}}.x dx\)
\(⇒I_1=x\sqrt{1-x^2}-{∫\sqrt{1-x^2} dx+∫\frac{-dx}{\sqrt{1-x^2}}}\)
\(⇒I_1=x\sqrt{1-x^2}-[{I_1+cos^{-1}x}]\)
\(⇒2I_1=x\sqrt{1-x^2}-cos^{-1}x\)
\(∴I_1=\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}cos^{-1}x\)
Substituting in equation(1),we obtain
\(I=\frac{xcos^{-1}x}{2}-\frac{1}{2}(\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}cos^{-1}x)-\frac{1}{2}cos^{-1}x\)
\(=\frac{(2x^2-1)}{4}cos^{-1}x-\frac{x}{4}\sqrt{1-x^2}+C\)