Evaluate the definite integral.?

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Answer 1 · 2017-11-16T20:21:19

$(2sqrt3-3)/3$

$int_0^(pi/6)sin(t)/cos^2(t)dt$

$=int_0^(pi/6)sin(t)/cos(t)*1/cos(t)dt$

$=int_0^(pi/6)tan(t)sec(t)dt$

$=[sec(t)]_0^(pi/6)$

$=sec(pi/6)-sec(0)$

$=2/sqrt3-1$

$=(2-sqrt3)/sqrt3$

$=(2sqrt3-3)/3$

Note that $d/dtsec(t)=sec(t)tan(t)$ (used above)