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)