# In the following exercises, find each indefinite integral, using appropriate

In the following exercises, find each indefinite integral, using appropriate substitutions.
$\int \frac{dx}{\sqrt{1-16{x}^{2}}}$
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Vivian Soares
Step 1
We need to evaluate the following integral
$I=\int \frac{dx}{\sqrt{1-16{x}^{2}}}$...(1)
Step 2
Let $x=\frac{\mathrm{sin}u}{4}⇒dx=\frac{\mathrm{cos}u}{4}du$ and substitute these in equation (1). We get,
$I=\int \frac{1}{\sqrt{1-16{\left(\frac{\mathrm{sin}u}{4}\right)}^{2}}}\left(\frac{\mathrm{cos}u}{4}du\right)$
$=\frac{1}{4}\int \frac{\mathrm{cos}u}{\sqrt{1-{\mathrm{sin}}^{2}u}}du$

$=\frac{1}{4}\int \frac{\mathrm{cos}u}{\mathrm{cos}u}du$
$=\frac{1}{4}\int du$
$=\frac{1}{4}u+C$

Hence, the integral is
$\int \frac{dx}{\sqrt{1-16{x}^{2}}}=\frac{1}{4}{\mathrm{sin}}^{-1}\left(4x\right)+C$