How do you evaluate # \int \frac { \cos x } { 1+ \sin ^ { 2} x } d x#?

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Answer 1 · 2018-07-08T12:41:39

#intcosx/(1+sin^2x)dx=arc tan (sinx)+c#

Here ,

#I=intcosx/(1+sin^2x)dx#

Subst. #color(blue)(sinx=u=>cosxdx=du#

#:.I=int1/(1+u^2)du#

#=>I=arc atan(u)+c#

Subst. back #color(blue)(u=sinx#

#I=arc tan (sinx)+c#

Answer 2 · 2018-07-08T12:43:50

The answer is #=arctan(sinx)+C#

Perform this integral by substitution

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

The integral is

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

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

Let #u=tantheta#, #=>#, #du=sec^2thetad theta#

#1+u^2=1+tan^2theta=sec^2theta#

Therefore,

#I=int(sec^2thetad theta)/(sec^2theta)#

#=intd theta#

#=theta#

#=arctan(u)#

#=arctan(sinx)+C#