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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