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CBSE Class 12-commerce Answered

integration of x(sqrt 1+xsquare)/sqrt 1- x square) within limits 0 and 1    
Asked by rajniyadav1977.ry | 25 Sep, 2019, 07:16: PM
answered-by-expert Expert Answer
integral subscript 0 superscript 1 fraction numerator straight x square root of 1 plus straight x squared end root over denominator square root of 1 minus straight x squared end root end fraction dx
put space straight x squared equals straight u
dx equals 1 half du
equals 1 half integral subscript 0 superscript 1 fraction numerator square root of 1 plus straight u end root over denominator square root of 1 minus straight u end root end fraction du
equals 1 half integral subscript 0 superscript 1 fraction numerator 1 plus straight u over denominator square root of 1 minus straight u squared end root end fraction du
equals 1 half integral subscript 0 superscript 1 fraction numerator 1 over denominator square root of 1 minus straight u squared end root end fraction du plus 1 half integral subscript 0 superscript 1 fraction numerator straight u over denominator square root of 1 minus straight u squared end root end fraction du
equals open square brackets 1 half space sin to the power of negative 1 end exponent straight u close square brackets subscript 0 superscript 1 plus 1 half integral subscript 0 superscript 1 fraction numerator straight u over denominator square root of 1 minus straight u squared end root end fraction du
now
1 half integral subscript 0 superscript 1 fraction numerator straight u over denominator square root of 1 minus straight u squared end root end fraction du
put
1 minus straight u squared equals straight t
du equals fraction numerator negative 1 over denominator 2 end fraction dt
equals fraction numerator negative 1 over denominator 4 end fraction integral subscript 1 superscript 0 fraction numerator 1 over denominator square root of straight t end fraction du
equals 1 fourth fraction numerator 1 over denominator begin display style 1 half end style end fraction open square brackets square root of straight t close square brackets subscript 0 superscript 1
equals open square brackets 1 half space sin to the power of negative 1 end exponent straight u close square brackets subscript 0 superscript 1 plus 1 half open square brackets square root of straight t close square brackets subscript 0 superscript 1
apply space limits
Answered by Arun | 26 Sep, 2019, 02:09: PM
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