Question #99aef

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Answer 1 · 2017-03-14T12:10:39

#0#

The integrand is an odd function that is defined on #[-pi/4, pi/4]#, so the integral is #0#

Answer 2 · 2017-03-15T16:01:39

Recognizing the function is odd is the quickest way to do this. We could, however, integrate using integration by parts.

For integration by parts, let:

#{(u=theta,=>,du=d theta),(dv=sec^2thetad theta,=>,v=tantheta):}#

#intthetasec^2thetad theta=thetatantheta-inttanthetad theta#

#=thetatantheta+int(-sintheta)/costhetad theta#

Letting #u=costheta# gives #du=-sinthetad theta#, so the remaining integral is in the form #int(du)/u=lnabsu#:

#=thetatantheta+lnabscostheta+C#

Then:

#int_(-pi/4)^(pi/4)thetasec^2thetad theta=[thetatantheta+lnabscostheta]_(-pi/4)^(pi/4)#

#=pi/4tan(pi/4)+lnabscos(pi/4)-(-pi/4tan(-pi/4)+lnabscos(-pi/4))#

#=pi/4+ln(1/sqrt2)-pi/4-ln(1/sqrt2)#

#=0#