`int (du)/(u sqrt(5 - u^2)) ` Evaluate the integral

`int(du)/(usqrt(5-u^2))`

` `

Let

`u=sqrt5sin(theta)`

`(du)/[d(theta)]=sqrt5cos(theta)`

`(du)=sqrt5cos(theta)d(theta)`

`int(du)/[usqrt(5-u^2)]`

`=int1/(sqrt5sin(theta))*[sqrt(5)cos(theta)d(theta)]/sqrt[5-(sqrt5sin(theta)^2)]`

`=int[cot(theta)d(theta)]/sqrt(5-5sin^2theta)`

`=int[cot(theta)d(theta)]/sqrt[5(1-sin^2theta)]`

`=int[cot(theta)d(theta)]/sqrt(5cos^2theta)`

`=int[cot(theta)d(theta)]/[sqrt(5)cos(theta)]`

`=int[cos(theta)d(theta)]/[sqrt(5)sin(theta)cos(theta)]`

`=int[d(theta)]/[sqrt(5)sin(theta)]`

`=int[csc(theta)d(theta)]/sqrt(5)`

`=1/sqrt(5)intcsc(theta)d(theta)`

`=1/sqrt(5)ln|sqrt(5)/u-sqrt(5-u^2)/u|+C`

`=1/sqrt(5)*ln|[sqrt5-sqrt(5-u^2)]/u|+C`

The final answer is 

`=1/sqrt(5)*ln|[sqrt5-sqrt(5-u^2)]/u|+C`