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What is the definite integral of #1/(36+x^2)# with bounds #[0, 6]#?

1 Answer

Jan 10, 2016

#pi/24#

Explanation:

There is a standard integral: #int(1/(a^2+x^2))dx = 1/atan^-1(x/a)+c#

So #int_0^6(1/(36+x^2))dx = [1/6tan^-1(x/6)]_0^6 #

#=(1/6tan^-1(1))-(1/6tan^-1(0)) = 1/6*pi/4 - 1/6*0 = pi/24#

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