How do you evaluate the definite integral #int (cosx)/(1+sinx)# from #[0,pi/2]#?

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Answer 1 · 2017-01-14T09:42:49

#ln2#

using the result

#int(f'(x))/(f(x))dx=ln|f(x)|+C#

since #d/(dx)(1+sinx)=cosx#

#int_0^(pi/2)(cosx)/(1+sinx)dx#

#=[ln|1+sinx|]_0^(pi/2)#

#=[ln(1+sin(pi/2))]-[cancel(ln(1+sin0))]#

#=ln2#