2021-12-30

Evaluate the integral.
$\int \frac{\mathrm{cos}x}{2+\mathrm{sin}x}$

Robert Pina

Step 1
Our Aim is to evaluate the integral given below:
$\int \frac{\mathrm{cos}\left(x\right)}{2+\mathrm{sin}\left(x\right)}dx$...(i)
Step 2
Considering the integral given by equation−(i), we have:−
$\int \frac{\mathrm{cos}\left(x\right)}{2+\mathrm{sin}\left(x\right)}dx$
For the integrand $\frac{\mathrm{cos}\left(x\right)}{2+\mathrm{sin}\left(x\right)}$, we will substitute $u=\mathrm{sin}\left(x\right)+2$ and $du=\mathrm{cos}\left(x\right)dx$
$=\int \frac{1}{u}du$
Since, the integral of $\frac{1}{u}$ is $\mathrm{log}\left(u\right)$
$=\mathrm{log}\left(u\right)$ +constant.
Substituting back for $u=\mathrm{sin}\left(x\right)+2$
Answer $⇒\mathrm{log}\left(\mathrm{sin}\left(x\right)+2\right)$ + constant.

Karen Robbins

Given that $\int \frac{\mathrm{cos}x}{2+\mathrm{sin}x}dx$
Let $2+\mathrm{sin}x=u$
$⇒\mathrm{cos}xdx=du$
Now $\int \frac{\mathrm{cos}x}{2+\mathrm{sin}x}dx=\int \frac{1}{u}du$
$=\mathrm{ln}|u|+c$
$\therefore \int \frac{\mathrm{cos}x}{2+\mathrm{sin}x}dx=\mathrm{ln}|2+\mathrm{sin}x|+c$

Vasquez

$\int \frac{\mathrm{cos}x}{2+\mathrm{sin}x}dx$
We put the expression $\mathrm{cos}\left(x\right)$ under the differential sign, i.e.:
$\mathrm{cos}\left(x\right)dx=d\left(\mathrm{sin}\left(x\right)\right),t=\mathrm{sin}\left(x\right)$
Then the original integral can be written as follows:
$\int \frac{1}{t+2}dt$
$\int \frac{1}{x+2}dx$
Calculate the table integral:
$\int \frac{1}{x+2}dx=\mathrm{ln}\left(x+2\right)$
$\mathrm{ln}\left(x+2\right)+c$
To write the final answer, it remains to substitute $\mathrm{sin}\left(x\right)$ instead of t.
$\mathrm{ln}\left(\mathrm{sin}\left(x\right)+2\right)+C$

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