Recall that
sin^2 (y/2) = (1 + cos(y))/2
so, √1 + cosy = √2 cos y/2
sin y = 2 sin y/2 cos y/2
So, the integrand becomes
2 siny/2 cosy/2 / √2cosy/2
= √2 sin(y/2)
Integral is just -2√2 cos(y/2)
ʃsin y dy/
√1 + cosy
please help..thanks..
that's integration of sin y dy divided by (over) square root of 1 + cosy.
3 answers
oops:
cos^2(y/2) = (1 + cos(y))/2
cos^2(y/2) = (1 + cos(y))/2
cos^2(y/2)=(1+2cos(y/2))/2
but 2 enter cos(y/2)
hence=(1+cos(2y/2))/2
=(1+cosy)/2
but 2 enter cos(y/2)
hence=(1+cos(2y/2))/2
=(1+cosy)/2