How do you find the integral of #cos(x) / sqrt(1+sin^2(x)#?

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Answer 1 · 2018-05-07T17:32:41

# ln|sinx+sqrt(1+sin^2x|)+C#.

If we subst. #sinx=t. :. cosxdx=dt#.

#:. intcosx/sqrt(1+sin^2x)dx#,

#=int1/sqrt(1+t^2)dt#,

#=ln|t+sqrt(1+t^2)|#.

But, #t=sinx#.

#:. I=ln|sinx+sqrt(1+sin^2x|)+C#.

Answer 2 · 2018-05-07T17:41:01

#I=ln|sinx+sqrt(1+sin^2x)|+c#

Here,

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

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

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

#=ln|u+sqrt(1+u^2)|+c#

Subst,back, #u=sinx#

#I=ln|sinx+sqrt(1+sin^2x)|+c#