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Question

Mathematics Question on integral

Integrate the function: xcos−1xxcos^{-1} x

Answer

The correct answer is: =(2x2−1)4cos−1x−x41−x2+C=\frac{(2x^2-1)}{4}cos^{-1}x-\frac{x}{4}\sqrt{1-x^2}+C
Let I=∫xcos−1x.dxI=∫xcos^{-1} x. dx Taking cos−1xcos^{−1} x as first function and xx as second function and integrating by parts,we
obtain
I=cos−1x∫x.dx−∫[(ddxcos−1x)∫xdx]dxI=cos^{-1}x∫x.dx-∫[{(\frac{d}{dx}cos^{-1}x)∫xdx}]dx
=cos−1x x22−∫−11−x2.x22dx=cos^{-1}x\, \frac{x^2}{2}-∫\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2} dx
=x2cos−12−12∫1−x2−11−x2dx=\frac{x^2cos^{-1}}{2}-\frac{1}{2}∫\frac{1-x^2-1}{\sqrt{1-x^2}} dx
=x2cos−12−12∫1−x2+(−11−x2)dx=\frac{x^2cos^{-1}}{2}-\frac{1}{2}\int{\sqrt{1-x^2}+(\frac{-1}{\sqrt{1-x^2}})}dx
=x2cos−12−12∫1−x2+12∫(−11−x2)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
=x2cos−12−12I1−12cos−1x...(1)=\frac{x^2cos^{-1}}{2}-\frac{1}{2}I_1-\frac{1}{2}cos^{-1}x...(1)
Where,I1=∫1−x2dxI_1=∫\sqrt{1-x^2} dx
⇒I1=x1−x2−∫ddx1−x2∫x.dx⇒I_1=x\sqrt{1-x^2}-∫\frac{d}{dx}\sqrt{1-x^2}∫x.dx
⇒I1=x1−x2−∫−2x21−x2.xdx⇒I_1=x\sqrt{1-x^2}-∫\frac{-2x}{2\sqrt{1-x^2}}.x dx
⇒I1=x1−x2−∫−x1−x2.xdx⇒I_1=x\sqrt{1-x^2}-∫\frac{-x}{\sqrt{1-x^2}}.x dx
⇒I1=x1−x2−∫1−x2−11−x2.xdx⇒I_1=x\sqrt{1-x^2}-∫\frac{1-x^2-1}{\sqrt{1-x^2}}.x dx
⇒I1=x1−x2−∫1−x2dx+∫−dx1−x2⇒I_1=x\sqrt{1-x^2}-{∫\sqrt{1-x^2} dx+∫\frac{-dx}{\sqrt{1-x^2}}}
⇒I1=x1−x2−[I1+cos−1x]⇒I_1=x\sqrt{1-x^2}-[{I_1+cos^{-1}x}]
⇒2I1=x1−x2−cos−1x⇒2I_1=x\sqrt{1-x^2}-cos^{-1}x
∴I1=x21−x2−12cos−1x∴I_1=\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}cos^{-1}x
Substituting in equation(1),we obtain
I=xcos−1x2−12(x21−x2−12cos−1x)−12cos−1xI=\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
=(2x2−1)4cos−1x−x41−x2+C=\frac{(2x^2-1)}{4}cos^{-1}x-\frac{x}{4}\sqrt{1-x^2}+C