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

Question

Emma Aldridge · Accepted Answer

We can use a trigonometric substitution to evaluate this integral.

Let #sintheta =2x # Then

#x=1/2sintheta #

distinguish

#dx=1/2costheta d theta #

Make a replacement

#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 #

Elijah Abram · Answer

The answer is: #1/2arcsin2x+c#. Recalling 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#.

Emma Aldridge · Answer

undefined To find the integral of $ \frac{1}{\sqrt{1-4x^2}} $ with respect to $ x $, you can use trigonometric substitution. Let $ x = \frac{1}{2}\sin(\theta) $. Then, $ dx = \frac{1}{2}\cos(\theta) d\theta $. Substitute these expressions into the integral and simplify. You should end up with $ \int \frac{1}{\sqrt{1-4x^2}} dx = \frac{1}{2} \int \frac{1}{\cos(\theta)} d\theta $. This simplifies further to $ \frac{1}{2} \int \sec(\theta) d\theta $. The integral of $ \sec(\theta) $ can be found using the natural logarithm function. Therefore, the final result is $ \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| + C $, where $ C $ is the constant of integration. Finally, substitute back $ \theta $ in terms of $ x $ to get the integral in terms of $ x $.