How do you evaluate the definite integral #intcosx/(1+sin^2x)dx# from #[0,1]#?

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Answer 1 · 2018-05-04T15:42:25

The answer is #=0.6995#

We need

#int(dx)/(1+x^2)=arctan(x)+ C#

First determine the definite integral by substitution

Let #u=sinx#, #=>#, #du=cosxdx#

Therefore,

#int(cosxdx)/(1+sin^2x)=int(du)/(u^2+1)#

#=arctan(u)#

#=arctan(sinx)+C#

The definite integral is

#int_0^1(cosxdx)/(1+sin^2x)#

#=[arctan(sinx)]_0^1#

#=(arctansin1)-(0)#

#=0.6995#