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:

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

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

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

$=0$