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How do you find the indefinite integral of #int 4/(1+t^2)dt#?

1 Answer

Jan 9, 2017

#int 4/(1+t^2)dt =4arctant+C#

Explanation:

Substitute:

#t=tanx#

#dt = (dx)/(cos^2x)#

#x=arctan t#

So:

#int 4/(1+t^2)dt = 4 int 1/(1+tan^2x) (dx)/(cos^2x)=4 int 1/(1+sin^2x/cos^2x) (dx)/(cos^2x)=4 int (dx)/(cos^2x+sin^2x) =4 int dx = 4x+C = 4arctant+C#

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