William Burnett

2022-01-03

Evaluate the integral.
$\int \frac{\mathrm{sin}hx}{1+\mathrm{cos}hx}dx$

peterpan7117i

Step 1
Given data:
The given integral is: $\int \frac{\mathrm{sin}hx}{1+\mathrm{cos}hx}dx$
Assume $1+\mathrm{cos}hx=u$ and differentiate it with respect to x.
$\left(0+\mathrm{sin}hx\right)dx=du$
$dx=\frac{du}{\mathrm{sin}hx}$
Step 2
Substitute the above-calculated values in the given expression.
$\int \frac{\mathrm{sin}hx}{1+\mathrm{cos}hx}dx=\int \frac{\mathrm{sin}hx}{u}\left(\frac{du}{\mathrm{sin}hx}\right)$
$=\int \frac{du}{u}$
$=\mathrm{ln}\left(u\right)+C$
Substitute $1+\mathrm{cos}hx$ for u in the above expression.
$\int \frac{\mathrm{sin}hx}{1+\mathrm{cos}hx}dx=\mathrm{ln}\left(1+\mathrm{cos}hx\right)+C$
Thus, the integral of the given expression is $\mathrm{ln}\left(1+\mathrm{cos}hx\right)+C$.

intacte87

$\int \frac{\mathrm{sin}h\left(x\right)}{1+\mathrm{cos}h\left(x\right)}dx$
$\int \frac{1}{t}dt$
Use $\int \frac{1}{x}dx=\mathrm{ln}\left(|x|\right)$ to evaluate the integral
$\mathrm{ln}\left(|t|\right)$
Substitute back $t=1+\mathrm{cos}h\left(x\right)$
$\mathrm{ln}\left(|1+\mathrm{cos}h\left(x\right)|\right)$
$\mathrm{ln}\left(|1+\mathrm{cos}h\left(x\right)|\right)+C$

Vasquez

Need to calculate:
$\begin{array}{}\int \frac{h\mathrm{sin}\left(x\right)}{h\mathrm{cos}\left(x\right)+1}dx\\ =-\int \frac{1}{u}du\\ \int \frac{1}{u}du\\ =\mathrm{ln}\left(u\right)\\ -\int \frac{1}{u}du\\ =-\mathrm{ln}\left(u\right)\\ =-\mathrm{ln}\left(h\mathrm{cos}\left(x\right)+1\right)\\ Answer:\\ =-\mathrm{ln}\left(|h\mathrm{cos}\left(x\right)+1|\right)+C\end{array}$

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