Question

if integral of 1/sqrt( 1 - y^2 ) is sin^(-1)y then what is 1/sqrt( 1 - 16y^2 ) ?? Medal for helping me :)

Answer 1

it's sin^-1(4y)
write the value which is at the place y^2 in term of square of that value then the term under square will come in answer as the argument of sin^-1...

Answer 2

thanks that's what I was looking for :)

Answer 3

you are welcome dear.........