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`int(dx)/(x^(2)sqrt(16-x^(2)))` has the value equal toA. `C-(1)/(4)tan^(-1)sec((x)/(4))`B. `(1)/(4)tan^(-1)sec((x)/(4))+C`C. `C-(sqrt(16-x^(2)))/(16x)`D. `(sqrt(16-x^(2)))/(16x)+C` |
Answer» Correct Answer - C `I=int(1)/(x^(2)sqrt(16-x^(2)))dx` Put `x=(1)/(t),` `dx=-(1)/(t^(2))dt therefore I=int(-(1)/(t^(2))dt)/((1)/(t)xx(1)/(t^(2))sqrt(16t^(2)-1))=int(-tdt)/(sqrt(16t^(2)-1))` Let `16t^(2)-1=u^(2), 32tdt = 2u du,` `tdt=(u)/(16)du thereforeI=-(1)/(16)int(udu)/(u)=-(u)/(16)+C=-(sqrt(16-x^(2)))/(16x)+C` |
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