What is the area enclosed by #r=thetacostheta-2sin(theta/2-pi) # for #theta in [pi/4,pi]#?

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Answer 1 · 2016-01-19T18:47:24

#A=int_(pi/4) ^pi 1/2 ( theta cos theta + 2 sin (theta/2))^2 d theta #

Area formula is #A =int_a^b 1/2 r^2 d theta#

Since #sin (theta/2 -pi) = - sin (pi - theta/2)= -sin (theta/2)#

Applying this formula #A=int_(pi/4) ^pi 1/2 ( theta cos theta + 2 sin (theta/2))^2 d theta #

This cannot be solved manually .