Question

Determine the following integral dx/√(a²-x²)³ by trigonometry substitution?

Answer 1

`int dx/(a^2-x^2)^(3/2)=1/a^2*x/sqrt(a^2-x^2)+C`

Explanation:

`int dx/(a^2-x^2)^(3/2)`

After using `x=asiny` and `dy=acosy*dy` transforms, this integral became

`int (acosy*dy)/(acosy)^3`

=`1/a^2int (secy)^2*dy`

=`1/a^2tany+C`

=`1/a^2siny/cosy+C`

After using `x=asiny`, `siny=x/a` and `cosy=sqrt(a^2-x^2)/a` inverse transforms, I found

`int dx/(a^2-x^2)^(3/2)`

=`1/a^2*(x/a)/(sqrt(a^2-x^2)/a)+C`

=`1/a^2*x/sqrt(a^2-x^2)+C`