CBSE Class 11-science Answered

$integral fraction numerator d x over denominator square root of a squared minus x squared end root end fraction equals sin to the power of negative 1 end exponent open parentheses x over a close parentheses$

prove this relation.

Asked by Bhavi Khatri | 18 Feb, 2016, 04:54: PM

Expert Answer

begin mathsize 11px style integral fraction numerator dx over denominator square root of straight a squared minus straight x squared end root end fraction Substitute space straight x equals asinθ... left parenthesis 1 right parenthesis rightwards double arrow dx equals acosθdθ The space integral space becomes space integral fraction numerator acosθdθ over denominator square root of straight a squared minus straight a squared sin squared straight theta end root end fraction equals integral fraction numerator acosθdθ over denominator square root of straight a squared left parenthesis 1 minus sin squared straight theta right parenthesis end root end fraction equals integral fraction numerator acosθdθ over denominator square root of straight a squared cos squared straight theta end root end fraction equals integral acosθdθ over acosθ equals integral dθ equals straight theta plus straight C Resubstituting space the space value space of space straight theta space from space left parenthesis 1 right parenthesis comma space we space get integral fraction numerator dx over denominator square root of straight a squared minus straight x squared end root end fraction equals sin to the power of negative 1 end exponent open parentheses straight x over straight a close parentheses plus straight C space left parenthesis Proved right parenthesis end style

Answered by satyajit samal | 19 Feb, 2016, 11:27: AM

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