1.

`int(sin^(-1)x)/((1-x^2)^(3//2))dx` का मान ज्ञात कीजिए ।

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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