Given ,

`int sqrt(16-4x^2)dx`

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

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

so here , For

`int sqrt(16-4x^2)dx -----(1)`

`x` can be given as

`x= sqrt(16/4) sin(u)= sqrt(4) sin(u) = 2sin(u)`

so,` x= 2sin(u)` => `dx = 2 cos(u) du`

Now substituting `x` in (1) we get,

`int sqrt(16-4x^2)dx `

=`int sqrt(16-4(2sin(u))^2) (2 cos(u) du)`

= `int sqrt(16-4*4(sin(u))^2) (2 cos(u) du)`

= `int sqrt(16-16(sin(u))^2) (2 cos(u) du)`

= `int sqrt(16(1-(sin(u))^2)) (2 cos(u) du)`

= `int sqrt(16(cos(u))^2) (2 cos(u) du)`

= `int (4cos(u)) (2 cos(u) du)`

=` int 8cos^2(u) du`

= `8 int cos^2(u) du`

= `8 int (1+cos(2u))/2 du`

= `(8/2) int (1+cos(2u)) du`

= `4 int (1+cos(2u)) du`

= `4 [int 1 du +int cos(2u) du]`

= `4 [u+(1/2)(sin(2u))] +c`

but `x= 2sin(u)`

=> `(x/2)= sin(u)`

=> `u= sin^(-1) (x/2)`

so,

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

=`4 [sin^(-1) (x/2)+1/2sin(2(sin^(-1) (x/2)))] +c`

so,

`int sqrt(16-4x^2)dx`

=`4sin^(-1) (x/2)+2sin(2(sin^(-1) (x/2))) +c `

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