Question #99aef

2 Answers
Mar 14, 2017

00

Explanation:

The integrand is an odd function that is defined on [-pi/4, pi/4][π4,π4], so the integral is 00

Mar 15, 2017

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

Explanation:

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