Question

find integral xsin inverse x/under root 1-x² dx​

Answer 1

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$∫1−x2xsin−1xdx[Here1stFunction=sin−1x2ndFunction=1−x2x] </p><p></p><p>⇒ Using integration by part 5</p><p></p><p>∫I.II dx=I ∫II dx−∫(dxdI∫II dx)dx</p><p></p><p>⇒ ∫1−x2xsin−1xdx=sin−1x∫1−x2xdx−∫(1−x21∫1−x2xdx)dx</p><p></p><p>Now Solving, Here ∫1−x2xdx=∫2t−dt Putting 1−x2=t</p><p></p><p>−2x dx=dt</p><p></p><p>x dx=2−dt</p><p></p><p>=2−11/2t=−t</p><p></p><p>=−1−x2</p><p></p><p>Hence, ∫1−x2xsin−1xdx=sin−1x[−1−x2]−∫1−x21×−1−x2dx</p><p></p><p>=−1−x2sin−1x+2x2+C where C= constant of Integration.</p><p></p><p>$

Answer 2

Answer:

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