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How do you evaluate the definite integral #int 4/sin^2(z) dz# from #[pi/6, pi/2]#?

1 Answer

Ratnaker Mehta · mason m

Sep 2, 2016

#4sqrt3#.

Explanation:

#int_(pi/6)^(pi/2) 4/sin^2 zdz=4int_(pi/6)^(pi/2)csc^2 zdz#

#=-4[cot z]_(pi/6)^(pi/2)#

#=-4[cot (pi/2)-cot (pi/6)]#

#=-4[0-sqrt3]#

#=4sqrt3#.

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