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`int sqrt(1-x) /sqrt(x) dx` Find the indefinite integral

Given,


`int sqrt(1-x)/sqrt(x) dx`


let us consider `u= sqrt(x)` ,


we can write it as `u^2 = x`


 Differentiating on both sides we get


=> `(2u)du = dx`


Now let us solve the integral ,


`int sqrt(1-x)/sqrt(x) dx`


=`int sqrt(1-u^2)/(u) ((2u)du)`   [as `x= u^2` ]


=`int 2*sqrt(1-u^2) du`


= `2int sqrt(1-u^2) du---(1)`


This can be solved by using the Trigonometric substitutions  (Trig substitutions)


For `sqrt(a-bx^2)` we have to take `x=` `sqrt(a/b) sin(v)`



so here , For


`2 int sqrt(1-u^2) du`   let us take `u = sqrt(1/1) sin(v) = sin(v)`


as `u= sin(v)` `=>` `du = cos(v) dv`


now substituting in (1) we get


`2int sqrt(1-u^2) du`


= `2int sqrt(1-(sin(v))^2) (cos(v) dv)`


= `2int sqrt((cos(v))^2) (cos(v) dv)`


= `2 int cos(v) cos(v) dv`


= `2 int cos^2(v) dv`


=`2 int (1+cos(2v))/2 dv`


= `(2/2) int (1+cos(2v)) dv`


= `int (1+cos(2v))dv`


=` [v+(1/2)(sin(2v))]+c`


but ,


`u = sin(v)`


=> `v= sin^(-1) u` and `u= sqrt(x)`


so,


`v= sin^(-1) (sqrt(x))`



now,


`v+1/2sin(2v)+c`


=`sin^(-1) (sqrt(x))+1/2sin(2sin^(-1) (sqrt(x)))+c`


so,


`int sqrt(1-x)/sqrt(x) dx`


=


=`sin^(-1) (sqrt(x))+1/2sin(2sin^(-1) (sqrt(x)))+c`

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