given to solve ,

`int sqrt(25-x^2) dx`

using the Trig Substitutions we can solve these type of integrals easily and the solution is as follows

for `sqrt(a-bx^2) ` we can take `x= sqrt(a/b) sin(u)`

so ,

`int sqrt(25-x^2) dx`

the `x = sqrt(25/1) sin(u) = 5sin(u)`

=> `dx = 5 cos(u) du`

so ,

`int sqrt(25-x^2) dx`

= `int sqrt(25-(5sin(u))^2) (5 cos(u) du)`

= `int sqrt(25-25(sin(u))^2) (5 cos(u) du)`

= `int 5 sqrt(1-sin^2 u )(5 cos(u) du)`

= `int 25 (cos(u))(cos(u)) du`

= `25 int cos^2(u) du`

= `25 int(1+cos(2u))/2 du`

= `(25/2) int (1+cos(2u)) du`

= `(25/2) [u+(1/2)(sin(2u))]+c`

but `x=5sin(u)`

=>` x/5 = sin(u)`

=> `u= arcsin(x/5)`

so,

`(25/2) [u+(1/2)(sin(2u))]+c`

=`(25/2) [(arcsin(x/5))+(1/2)(sin(2(arcsin(x/5))))]+c`