Evaluate integral of 1/( square root of 1-16x^2) with respect to x

$\int \frac{1}{\sqrt{1 - 16 x^{2}}} d x$

Write the expression using exponents.

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Rewrite $1$ as $1^{2}$ .

$\int \frac{1}{\sqrt{1^{2} - 16 x^{2}}} d x$

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

$\int \frac{1}{\sqrt{1^{2} - {(4 x)}^{2}}} d x$

Let $u = 4 x$ . Then $d u = 4 d x$ , so $\frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = 4 x$ . Find $\frac{d u}{d x}$ .

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Rewrite.

$\frac{4}{1}$

Divide $4$ by $1$ .

$4$

Rewrite the problem using $u$ and $d u$ .

$\int \frac{1}{\sqrt{1^{2} - u^{2}}} \cdot \frac{1}{4} d u$

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{4} \int \frac{1}{\sqrt{1^{2} - u^{2}}} d u$

The integral of $\frac{1}{\sqrt{1^{2} - u^{2}}}$ with respect to $u$ is $\arcsin (u)$

$\frac{1}{4} (\arcsin (u) + C)$

Simplify.

$\frac{1}{4} \arcsin (u) + C$

Replace all occurrences of $u$ with $4 x$ .

$\frac{1}{4} \arcsin (4 x) + C$

Evaluate integral of 1/( square root of 1-16x^2) with respect to x

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