How do you integrate $\int \frac{1}{\sqrt{1 - 4 x^{2}}}$ $int 1/sqrt(1-4x^2)$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{1 - 4 x^{2}}}$ $int 1/sqrt(1-4x^2)$ by trigonometric substitution?

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Answer 1 · 2016-10-12T22:00:33

$\int \frac{1}{\sqrt{1 - 4 x^{2}}} d x = \frac{1}{2} {sin}^{- 1} (2 x) + C$ $int 1/(sqrt(1-4x^2)) dx = 1/2 sin^(-1) (2x) + C$

Explanation:

Let $x = \frac{1}{2} sin θ$ $x = 1/2 sin theta$

Then:

$\int \frac{1}{\sqrt{1 - 4 x^{2}}} d x = \int \frac{1}{\sqrt{1 - {sin}^{2} θ}} \frac{d \frac{1}{2} sin θ}{d θ} d θ$ $int 1/(sqrt(1-4x^2)) dx = int 1/sqrt(1 - sin^2 theta) (d 1/2 sin theta)/(d theta) d theta$

$\int \frac{1}{\sqrt{1 - 4 x^{2}}} d x = \int \frac{\frac{1}{2} cos θ}{cos θ} d θ$ $color(white)(int 1/(sqrt(1-4x^2)) dx) = int (1/2 cos theta)/cos theta d theta$

$\int \frac{1}{\sqrt{1 - 4 x^{2}}} d x = \int \frac{1}{2} d θ$ $color(white)(int 1/(sqrt(1-4x^2)) dx) = int 1/2 d theta$

$\int \frac{1}{\sqrt{1 - 4 x^{2}}} d x = \frac{1}{2} θ + C$ $color(white)(int 1/(sqrt(1-4x^2)) dx) = 1/2 theta + C$

$\int \frac{1}{\sqrt{1 - 4 x^{2}}} d x = \frac{1}{2} {sin}^{- 1} (2 x) + C$ $color(white)(int 1/(sqrt(1-4x^2)) dx) = 1/2 sin^(-1) (2x) + C$

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How do you integrate $\int \frac{1}{\sqrt{1 - 4 x^{2}}}$ $int 1/sqrt(1-4x^2)$ by trigonometric substitution?

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Explanation:

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How do you integrate ∫1√1−4x2int 1/sqrt(1-4x^2) by trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{1 - 4 x^{2}}}$ $int 1/sqrt(1-4x^2)$ by trigonometric substitution?