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...

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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`