Question

Evaluate : `int_0^3dx/(9+x^2)`

Answer 1

Given,

`I=int_0^3dx/(9+x^2)=int_0^3dx/(3^2+x^2)`

We know that, `intdx/(x^2+a^2)=1/atan^(-1)(x/a)+C`

Therefore,

`I=int_0^3dx/(x^2+3^2)`

`=1/3[tan^(-1)(x/3)]_0^3`

`=1/3[tan^(-1)1-tan^(-1)0]`

`=1/3[pi/4-0]`

`=pi/12`