Indefinite Integral: $\int \frac{1}{\sqrt{\sin 2x}}dx$

MHB
Thread starter juantheron
Start date Apr 15, 2012
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Indefinite Indefinite integral Integral

In summary, an indefinite integral is an antiderivative of a function that reverses the process of differentiation. To solve indefinite integrals, various techniques such as substitution and integration by parts can be used, along with basic rules of integration. The general formula for indefinite integrals is ∫f(x)dx = F(x) + C, and to integrate a function with a square root in the denominator, substitution and trigonometric substitutions may be necessary. The indefinite integral of $\frac{1}{\sqrt{\sin 2x}}$ is $\frac{1}{\sqrt{2}}\arctan \left( \sqrt{\frac{\sin 2x}{\cos 2x}} \right) + C

Apr 15, 2012

juantheron

$\displaystyle \int \frac{1}{\sqrt{\sin 2x}}dx$

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Apr 15, 2012

CaptainBlack

jacks said:

$\displaystyle \int \frac{1}{\sqrt{\sin 2x}}dx$

How are you expecting this to be expressed? It is not elementary, but is an elliptic integral of the first kind.

CB

Last edited: Apr 15, 2012

1. What is an indefinite integral?

An indefinite integral is an antiderivative of a function. It is a mathematical operation that reverses the process of differentiation, and gives a family of functions that differ by a constant.

2. How do you solve indefinite integrals?

To solve an indefinite integral, you need to use the techniques of integration, such as substitution, integration by parts, or trigonometric substitutions. You also need to know the basic rules of integration, such as the power rule and the constant multiple rule.

3. What is the general formula for indefinite integrals?

The general formula for indefinite integrals is ∫f(x)dx = F(x) + C, where F(x) is the antiderivative of f(x) and C is the constant of integration.

4. How do you integrate a function with a square root in the denominator?

To integrate a function with a square root in the denominator, you can use the substitution method, where you replace the square root with a new variable and then solve the resulting integral. In some cases, you may also need to use trigonometric substitutions.

5. What is the indefinite integral of $\frac{1}{\sqrt{\sin 2x}}$?

The indefinite integral of $\frac{1}{\sqrt{\sin 2x}}$ is given by $\int \frac{1}{\sqrt{\sin 2x}}dx = \frac{1}{\sqrt{2}}\arctan \left( \sqrt{\frac{\sin 2x}{\cos 2x}} \right) + C$, where C is the constant of integration.