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

Answer 1

Aurora Alvarado

#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#

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Answer 2

Emma Aldridge

# 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#.

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Answer 3

Charles Arocha

To find the integral of ( \frac{\cos(x)}{\sqrt{1+\sin^2(x)}} ), we can use the substitution method.

Let ( u = \sin(x) ), then ( du = \cos(x) dx ).

Now, the integral becomes ( \int \frac{1}{\sqrt{1+u^2}} du ), which is the integral of ( \frac{1}{\sqrt{1+x^2}} ), a standard integral.

This integral is equivalent to ( \sinh^{-1}(x) + C ), where ( \sinh^{-1}(x) ) represents the inverse hyperbolic sine function.

Substituting back ( u = \sin(x) ), the final result is ( \sinh^{-1}(\sin(x)) + C ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.