How do you find #\int _ { \pi / 2} ^ { 2\pi } \sqrt { 1- \cos ^ { 2} t } d t#?

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Answer 1 · 2016-12-12T00:32:33

#int_(pi/2)^(2pi)(sqrt(1-cos^2t))dt#

#=int_(pi/2)^(2pi)(sqrt(sin^2t))dt#

#=int_(pi/2)^(2pi)(abs(sint))dt#

#=int_(pi/2)^(pi)(sint)dt+int_(pi)^(2pi)(-sint)dt#

#=[-cost]_(pi/2)^(pi)#

#+[cost]_(pi)^(2pi)#

#=[(cospi)-(cos(pi/2))]+[(cos2pi)-(cospi)]#

#=[-1-0]+[1--1]#

#=-1+2#

#=1#