- #1
XtremePhysX
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Homework Statement
Find:
Homework Equations
[tex]\int \sqrt{\frac{x}{1-x}}dx[/tex]
The Attempt at a Solution
I tried to use u substitution with u=1-x but it did work.
XtremePhysX said:Homework Statement
Find:
Homework Equations
[tex]\int \sqrt{\frac{x}{1-x}}dx[/tex]
The Attempt at a Solution
I tried to use u substitution with u=1-x but it did work.
XtremePhysX said:bump :)
XtremePhysX said:[tex]sin^{-1}\sqrt{x}-\frac{2sin(sin^{-1}\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(sin^{-1}\sqrt{x})}{2}=sin^{-1}\sqrt{x}-\frac{2(\sqrt{x})cos(\sqrt{1-x})}{2}[/tex]
Is this right?
XtremePhysX said:bump :)
im really sorry, I am new here. Please excuse my actions.
where can i find the rules to read them?
Solving an integral involves finding the antiderivative of a given function and evaluating it at the upper and lower limits of integration. This process is also known as integration or antidifferentiation.
A definite integral has specific limits of integration and yields a single numerical value, while an indefinite integral does not have limits and results in a function.
Substitution is a technique used to simplify an integral by replacing a variable with a new one. This new variable can then be integrated more easily, and the answer can be converted back to the original variable.
No, integration by parts can only be used for certain types of integrals, such as those that involve products of functions. It is not a universal method for solving all integrals.
The choice of integration technique depends on the form of the integrand and the given limits of integration. It is important to be familiar with a variety of integration methods and to choose the most appropriate one for each problem.