Harold Kessler

2021-12-28

Solve the integral.
$\int \frac{\mathrm{cos}x}{1+{\mathrm{sin}}^{2}x}dx$

Tiefdruckot

Step 1
The given integral is:
$\int \frac{\mathrm{cos}x}{1+{\mathrm{sin}}^{2}x}dx$
To solve this integral,
Consider, $\mathrm{sin}x=t$..(1)
Now, differentiate equation (1) on both sides.
$\mathrm{cos}xdx=dt$...(2)
From equations (1) and (2) plug these values in the given integral.
$\int \frac{\mathrm{cos}x}{1+{\mathrm{sin}}^{2}x}dx=\int \left(\frac{1}{1+{t}^{2}}\right)dt$
Step 2
Now, integrate further,
$\int \frac{\mathrm{cos}x}{1+{\mathrm{sin}}^{2}x}dx={\mathrm{tan}}^{-1}\left(t\right)+c$
Now, from equation (1), again replace t with $\mathrm{sin}\left(x\right)$.
That is,
$\int \frac{\mathrm{cos}x}{1+{\mathrm{sin}}^{2}x}dx={\mathrm{tan}}^{-1}\left(\mathrm{sin}x\right)+c$
$\int \frac{\mathrm{cos}x}{1+{\mathrm{sin}}^{2}x}dx={\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{x}{\sqrt{1-{x}^{2}}}\right)\right)+c$
$\int \frac{\mathrm{cos}x}{1+{\mathrm{sin}}^{2}x}dx=\frac{x}{\sqrt{1-{x}^{2}}}+c$
Where c is an arbitrary constant.

Jack Maxson

$\int \frac{\mathrm{cos}\left(x\right)}{1+{\mathrm{sin}\left(x\right)}^{2}}dx$
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:
$\int \frac{1}{{t}^{2}+1}dt$
This is a tabular integral:
$\int \frac{1}{{t}^{2}+1}dt=\mathrm{arctan}\left(t\right)+C$
To write the final The answer is, it remains to substitute $\mathrm{sin}\left(x\right)$ instead of t.
$\mathrm{arctan}\left(\mathrm{sin}\left(x\right)\right)+C$

Vasquez

$\begin{array}{}\int \frac{\mathrm{cos}\left(x\right)}{1+\mathrm{sin}\left(x{\right)}^{2}}dx\\ \int \frac{1}{1+{t}^{2}}dt\\ \frac{1}{1}×\mathrm{arctan}\left(\frac{t}{1}\right)\\ \frac{1}{1}×\mathrm{arctan}\left(\frac{\mathrm{sin}\left(x\right)}{1}\right)\\ \mathrm{arctan}\left(\mathrm{sin}\left(x\right)\right)\\ Answer:\\ \mathrm{arctan}\left(\mathrm{sin}\left(x\right)\right)+C\end{array}$