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## Homework Statement

Find the area under bounded by y = cos x and y = sin 2x on the intervacl [0, pi/2]

## The Attempt at a Solution

cos x = sin 2x

2sin x cos x - cos x = 0

x = pi/6 and pi/2

shouldn't the intergral be: [ S(0,pi/6) cos x - sin 2x ] - [ S(pi/6,pi/2) sin 2x - cos x ]

I have [ S(0,pi/6) cos x - sin 2x ] - [ S(pi/6,pi/2) sin 2x + cos x ]

why is it sin 2x + cos x?