What is the integral of # [ (1+cos^2(x)) / sin(x) ] dx#?

Answer 1

#= -2 ln | csc x - cot x| + cos x + C#

#int (1+cos^2(x)) / sin(x) dx#

is just crying out for the application of the Pythagorean identity

#=int (1+ color(red)(1 - sin^2 x)) / sin(x) dx#
#=int (2 - sin^2 x) / sin(x) dx#
#=int 2 csc x - sin x dx#

these are well known integrals

#= -2 ln | csc x - cot x| + cos x + C#
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Answer 2

To integrate ( \frac{{1+\cos^2(x)}}{{\sin(x)}} ) with respect to ( x ), you can use trigonometric identities to simplify the expression. First, rewrite ( \cos^2(x) ) as ( 1 - \sin^2(x) ). Then, you'll have:

[ \frac{{1+\cos^2(x)}}{{\sin(x)}} = \frac{{1+(1-\sin^2(x))}}{{\sin(x)}} = \frac{{2 - \sin^2(x)}}{{\sin(x)}} ]

Now, let ( u = \sin(x) ), then ( du = \cos(x)dx ). So, the integral becomes:

[ \int \frac{{2 - u^2}}{{u}} du ]

This can be split into two separate integrals:

[ \int \frac{{2}}{{u}} du - \int \frac{{u^2}}{{u}} du ]

[ = 2\ln|u| - \int u du ]

[ = 2\ln|\sin(x)| - \frac{{\sin^2(x)}}{{2}} + C ]

Where ( C ) is the constant of integration.

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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.

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.

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.

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.

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