Deragz

2021-12-17

Find the indefinite integral.
$\int \frac{-1}{\sqrt{1-{\left(4t+1\right)}^{2}}}dt$

Wendy Boykin

Step 1
We have the given integral,
$\int \frac{-1}{\sqrt{1-{\left(4t+1\right)}^{2}}}dt$
By substituting u=4t+1, that implies du=4dt
Therefore, the integral becomes,
$=\int \frac{-1}{\sqrt{1-{u}^{2}}}\left(\frac{1}{4}\right)du$
$=\frac{-1}{4}\int \frac{1}{\sqrt{1-{u}^{2}}}du$
Step 2
Now, by substituting $\mathrm{sin}x=u$, that implies, $\mathrm{cos}xdx=du$
Therefore, the integral becomes,
$=\frac{-1}{4}\int \frac{1}{\sqrt{1-{\mathrm{sin}}^{2}x}}\left(\mathrm{cos}x\right)dx$
$=\frac{-1}{4}\int \frac{1}{\sqrt{{\mathrm{cos}}^{2}x}}\left(\mathrm{cos}x\right)dx$
$=\frac{-1}{4}\int \frac{1}{\mathrm{cos}x}\left(\mathrm{cos}x\right)dx$
$=\frac{-1}{4}\int dx$
$=\frac{-1}{4}x+C$
Step 3
But we know, $\mathrm{sin}x=u$, therefore $x={\mathrm{sin}}^{-1}\left(u\right)$
By substituting this value of x,
$=-\frac{1}{4}{\mathrm{sin}}^{-1}u+C$
Then by resubstituting u=4t+1,
$=-\frac{1}{4}{\mathrm{sin}}^{-1}\left(4t+1\right)+C$

Tiefdruckot

$\int \frac{-1}{\sqrt{1-{\left(4t+1\right)}^{2}}}dt$
$\int -\frac{1}{4\sqrt{1-{u}^{2}}}du$
$-\frac{1}{4}\cdot \int \frac{1}{\sqrt{1-{u}^{2}}}du$
$-\frac{1}{4}\cdot \mathrm{arcsin}\left(u\right)$
$-\frac{1}{4}\cdot \mathrm{arcsin}\left(4t+1\right)$
$-\frac{\mathrm{arcsin}\left(4t+1\right)}{4}$
Solution:
$-\frac{\mathrm{arcsin}\left(4t+1\right)}{4}+C$

nick1337

Step 1
Given:
$\int -\frac{1}{\sqrt{1-\left(4t+1{\right)}^{2}}}dt$
Substitution $u=4t+1⇒\frac{du}{dt}=4$
$=-\frac{1}{4}\int \frac{1}{\sqrt{1-{u}^{2}}}du$
This is the well-known tabular integral:
$=\mathrm{arcsin}\left(u\right)$
We substitute the already calculated integrals:
$-\frac{1}{4}\int \frac{1}{\sqrt{1-{u}^{2}}}du$
$=-\frac{\mathrm{arcsin}\left(u\right)}{4}$
Reverse replacement u=4t+1:
$=-\frac{\mathrm{arcsin}\left(4t+1\right)}{4}$
$=-\frac{\mathrm{arcsin}\left(4t+1\right)}{4}+C$