Integration of y upon root 1-y²
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Integration of y(dy)/(1-y^2)^0.5
Let 1-y^2=x
=>-2y(dy)=dx
Therefore, integration of -(dx)/2x^0.5
=integration of -x^(-0.5) (dx)/2
=(-x^(-0.5+1))/2(-0.5+1)
=(-x^0.5)/(2*0.5)
=-x^0.5
Putting value of x we get
=-(1-y^2)^0.5
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Let 1-y^2=x
=>-2y(dy)=dx
Therefore, integration of -(dx)/2x^0.5
=integration of -x^(-0.5) (dx)/2
=(-x^(-0.5+1))/2(-0.5+1)
=(-x^0.5)/(2*0.5)
=-x^0.5
Putting value of x we get
=-(1-y^2)^0.5
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