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

1 Answer
Apr 23, 2016

#-lnabs(1+cosx)+C#

Explanation:

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#