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Question
Integrate the function xcos−1x
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Solution
Let I=∫xcos−1xdxApplying integration by parts, by taking cos−1x as first function and x as second function
I=cos−1x∫xdx−∫{(ddxcos−1x)}∫(xdx)dx
=cos−1xx22−∫−1√1−x2⋅x22dx
=x2cos−1x2−12∫1−x2−1√1−x2dx
=x2cos−1x2−12∫{√1−x2+(−1√1−x2)}dx
=x2cos−1x2−12∫√1−x2dx−12∫(−1√1−x2)dx
=x2cos−1x2−12I1−12cos−1x .......(1)
where, I1=∫√1−x2dxApplying integration by parts
⇒I1=√1−x2∫1dx−∫ddx√1−x2∫1dx
⇒I1=x√1−x2−∫−x2√1−x2⋅dx
⇒I1=x√1−x2−∫−x2√1−x2dx
⇒I1=x√1−x2−∫1−x2−1√1−x2dx
⇒I1=x√1−x2−{∫√1−x2dx+∫−dx√1−x2}
⇒I1=x√1−x2−{I1+cos−1x}
⇒2I1=x√1−x2−cos−1x
I1=x2√1−x2−12cos−1x
Substituting in (1), we obtain
I=x2cos−1x2−12(x2√1−x2−12cos−1x)−12cos−1x
=(2x2−1)4cos−1x−x4√1−x2+C
Applying integration by parts, by taking cos−1x as first function and x as second function
I=cos−1x∫xdx−∫{(ddxcos−1x)}∫(xdx)dx
I=cos−1x∫xdx−∫{(ddxcos−1x)}∫(xdx)dx
=cos−1xx22−∫−1√1−x2⋅x22dx
=x2cos−1x2−12∫1−x2−1√1−x2dx
=x2cos−1x2−12∫{√1−x2+(−1√1−x2)}dx
=x2cos−1x2−12∫√1−x2dx−12∫(−1√1−x2)dx
=x2cos−1x2−12I1−12cos−1x .......(1)
where, I1=∫√1−x2dx
Applying integration by parts
⇒I1=√1−x2∫1dx−∫ddx√1−x2∫1dx
⇒I1=x√1−x2−∫−x2√1−x2⋅dx
⇒I1=x√1−x2−∫−x2√1−x2dx
⇒I1=x√1−x2−∫1−x2−1√1−x2dx
⇒I1=x√1−x2−{∫√1−x2dx+∫−dx√1−x2}
⇒I1=x√1−x2−{I1+cos−1x}
⇒2I1=x√1−x2−cos−1x
I1=x2√1−x2−12cos−1x
Substituting in (1), we obtain
I=x2cos−1x2−12(x2√1−x2−12cos−1x)−12cos−1x
=(2x2−1)4cos−1x−x4√1−x2+C
⇒I1=√1−x2∫1dx−∫ddx√1−x2∫1dx
⇒I1=x√1−x2−∫−x2√1−x2⋅dx
⇒I1=x√1−x2−∫−x2√1−x2dx
⇒I1=x√1−x2−∫1−x2−1√1−x2dx
⇒I1=x√1−x2−{∫√1−x2dx+∫−dx√1−x2}
⇒I1=x√1−x2−{I1+cos−1x}
⇒2I1=x√1−x2−cos−1x
I1=x2√1−x2−12cos−1x
Substituting in (1), we obtain
I=x2cos−1x2−12(x2√1−x2−12cos−1x)−12cos−1x
=(2x2−1)4cos−1x−x4√1−x2+C
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