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What is the antiderivative of # sinx/(1+cosx)#?

Answer 1

#-lnabs(1+cosx)+C#

We want to find

#intsinx/(1+cosx)dx#

To do so, we will use substitution: let #u=1+cosx#. This implies that #du=-sinxdx#.

To attain the #du=-sinxdx# value in the integral, multiply the interior of the integral by #-1#. Balance this by multiplying the exterior by #-1# as well.

#=-int(-sinx)/(1+cosx)dx#

Substituting in our #u# and #du# values, this gives us

#=-int1/udu#

This is a common integral:

#=-lnabsu+C#

Using our value for #u#:

#=-lnabs(1+cosx)+C#

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Answer 2

Cameron Alexander

The antiderivative of ( \frac{\sin(x)}{1+\cos(x)} ) is ( -\ln|1+\cos(x)| + 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.