Question

Find 11tanxdx.\int_{-1}^{1} \tan x d x.

Solution

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By Property 5, we can rewrite the integral given as

=10tanxdx+01tanxdx= \int_{-1}^{0}\tan x\,dx + \int_{0}^{1}\tan x\,dx

However, tanx\tan x is an odd function. That is, for every xx in its domain, f(x)=f(x)f(-x) = -f(x). Thus, the graph sections corresponding to the two terms above are symmetrical:

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