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what is the integration of (sin-1x)2?
?(sin-1x)2dx put sin-1x=t => x=sint dx=cost dt ?t2costdt =t2sint-?2t.sintdt =t2sint-2[-tcost+?costdt] =t2sint+2tcost-2sint +C =(sin-1x)2x+2sin-1xv(1-x2)-2x+ C Thanks & Regards Rinkoo Gupta AskIITians Faculty
?(sin-1x)2dx
put sin-1x=t => x=sint
dx=cost dt
?t2costdt
=t2sint-?2t.sintdt
=t2sint-2[-tcost+?costdt]
=t2sint+2tcost-2sint +C
=(sin-1x)2x+2sin-1xv(1-x2)-2x+ C
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
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