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Question
Evaluate : `int_0^3dx/(9+x^2)`
Solution
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`
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