Can you think of a different way?

Question

Any way you can calculate this integral
#int_0^(pi/4)(sqrt(1+cosx)) dx#
without using the trigonometric identity #1+cosx=2cos^2(x/2)# ?

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Answer 1 · 2018-02-02T15:06:23

Please see below.

#sqrt(1+cosx) = sqrt(1+cosx) * sqrt(1-cosx)/sqrt(1-cosx)#

# = sqrt(sin^2x)/sqrt(1-cosx)#

# = abs(sinx)/sqrt(1-cosx)#

So,

#int_0^(pi/4) sqrt(1+cosx) dx = int_0^(pi/4) (sinx)/sqrt(1-cosx) dx#

Now use #u = 1-cosx#