How do you integrate #tanx/(1+cosx)#?

Answer 1
#inttanx/(1+cosx)dx=intsinx/(cosx(1+cosx))dx#

let

#u=cosx# #du=-sinxdx#

partial fractions

#-int(du)/(u(u+1))=-int(1/u-1/(u+1))du#
#-int(1/u-1/(u+1))du=int(du)/(u+1)-int(du)/u#
#-int(du)/u=-ln|u|#
For #int(du)/(u+1)#
set #w=u+1# #dw=du#
#int(dw)/w = ln|w|#

Plug back in for u and x

#ln|w|= ln|u+1|#

So far we have

#ln|u+1|-ln|u|#
Remember #u=cosx#
#ln|cosx+1|-ln|cosx|#
And #+ C#
#inttanx/(1+cosx)dx=ln|cosx+1|-ln|cosx|+C#
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Answer 2

To integrate tan(x)/(1+cos(x)), you can use the trigonometric substitution method. Let ( u = \tan(\frac{x}{2}) ). Then, ( \cos(x) = \frac{1-u^2}{1+u^2} ) and ( \sin(x) = \frac{2u}{1+u^2} ). Substitute these into the integral, and you'll end up with an integral involving ( u ). After integrating with respect to ( u ), you can then substitute back in terms of ( x ) to get the final result.

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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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