∫xsin−1⁡xdx is equal to

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sin−1⁡x22x2−1+14x1−x2+C

sin−1⁡x42x2−1+14x1−x2+C

sin−1⁡x42x2−1+12x1−x2+C

None of the above

answer is B.

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Detailed Solution

Let I=∫x sin−1⁡xdxOn taking sin-1 x as first function and x as second function and integrating by parts, we get[∵ inverse functions comes before,algebraic functions in ILATE]I=sin−1⁡x∫xdx−∫ddxsin−1⁡x∫xdxdx =sin−1⁡x⋅x22−∫11−x2⋅x22dx =x22⋅sin−1⁡x−∫1−1−x21−x2⋅12dx[add and subtract 1 in numerator of second term]=x22⋅sin−1⁡x−12∫11−x2dx+12∫1−x21−x2dx=x22⋅sin−1⁡x−12sin−1⁡x+12∫1−x2dx⇒ I=x22⋅sin−1⁡x−12sin−1⁡x+1212x1−x2+12sin−1⁡x+C⇒ I=sin−1⁡xx22−14+14x1−x2+C⇒ I=sin−1⁡x42x2−1+14x1−x2+C

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