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`int(sin^(-1)x)/((1-x^2)^(3//2))dx` का मान ज्ञात कीजिए । |
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Answer» दिया है - `int(sin^(-1)x)/((1-x^2)^(3//2))dx` यदि `x=sintrArr dx=cost dt " व " t =sin^(-1)x`, तब `int(sin^(-1)x)/((1-x^2)^(3//2)dx=int(t cost)/((1-sin^2t)^(3//2)dt` `=int(tcost)/(cos^3t)dt =intt sec^2tdt` खण्डश: समाकलन करने पर `=t(tant)-int1 tantdt` `=t (tant)+log(cost)+c` `=(sin^(-1)x)x/sqrt(1-x^2)+log(sqrt(1-x^2))+c` यहाँ `cost=sqrt(1-x^2)" व " tant =x/sqrt(1-x^2)` `= (x(sin^(-1)x))/sqrt(1-x^2)+1/2log(1-x^2)+c` |
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