How do you find the integral of $1/sqrt(1-4x^2)dx$ ?

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Answer 1 · 2015-02-28T21:07:29

We can evaluate this integral by using a trigonometric substitution.

Let $sintheta =2x$ Then

$x=1/2sintheta$

differentiate

$dx=1/2costheta d theta$

Make substitution

$1/2intcostheta/(sqrt(1-4(1/2sintheta)^2))d theta$

$1/2intcostheta/(sqrt(1-4(1/4sin^2theta)))d theta$

$1/2intcostheta/(sqrt(1-sin^2theta))d theta$

$1/2intcostheta/(sqrt(cos^2theta))d theta$

$1/2intcostheta/(costheta)d theta$

$1/2intd theta$

Integrating we have $1/2theta$

Defining $theta$ in terms of $x$

$theta=arcsin(2x)$ Therefore

$1/2arcsin(2x)+C$

Answer 2 · 2015-02-28T21:51:49

The answer is: $1/2arcsin2x+c$ .

Remembering that:

$int(f'(x))/sqrt(1-[f(x)]^2)dx=arcsinf(x)+c$ ,

so:

$int1/sqrt(1-4x^2)dx=int1/sqrt(1-(2x)^2)dx=1/2int2/(sqrt(1-(2x)^2))dx=$

$=1/2arcsin2x+c$ .

How do you find the integral of 1/sqrt(1-4x^2)dx1/sqrt(1-4x^2)dx?