What is the antiderivative of $\frac{sin x}{1 + cos x}$ $sinx/(1+cosx)$ ?

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What is the antiderivative of $\frac{sin x}{1 + cos x}$ $sinx/(1+cosx)$ ?

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Answer 1 · 2016-04-23T17:44:35

$- ln | 1 + cos x | + C$ $-lnabs(1+cosx)+C$

Explanation:

We want to find

$\int \frac{sin x}{1 + cos x} d x$ $intsinx/(1+cosx)dx$

To do so, we will use substitution: let $u = 1 + cos x$ $u=1+cosx$ . This implies that $d u = - sin x d x$ $du=-sinxdx$ .

To attain the $d u = - sin x d x$ $du=-sinxdx$ value in the integral, multiply the interior of the integral by $- 1$ $-1$ . Balance this by multiplying the exterior by $- 1$ $-1$ as well.

$= - \int \frac{- sin x}{1 + cos x} d x$ $=-int(-sinx)/(1+cosx)dx$

Substituting in our $u$ $u$ and $d u$ $du$ values, this gives us

$= - \int \frac{1}{u} d u$ $=-int1/udu$

This is a common integral:

$= - ln | u | + C$ $=-lnabsu+C$

Using our value for $u$ $u$ :

$= - ln | 1 + cos x | + C$ $=-lnabs(1+cosx)+C$

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What is the antiderivative of $\frac{sin x}{1 + cos x}$ $sinx/(1+cosx)$ ?

1 Answer

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