- #1
vincentchan
- 609
- 0
anyone has an easy way to do the following problem..
[tex] \int \sqrt{\tan x} dx [/tex]
[tex] \int \sqrt{\tan x} dx [/tex]
If [itex]du = \sec^2x dx[/itex] then:mathwonk said:the basic idea of substitution is to try to put a new symbol in for anything you find confusing. Like here you might try u = tan(x), and then of course you have to put du
= sec^2(x)dx. See if that makes it any better. you may to do some more substitutions later.
The process for solving this calculus problem is called integration. Integration is the inverse operation of differentiation, which is used to find the area under a curve. In this case, we are integrating the function √tan(x) with respect to the variable x.
When approaching this type of integration problem, it is helpful to use trigonometric identities to rewrite the function in a more manageable form. In this case, we can use the identity tan(x) = sin(x)/cos(x) to rewrite the function as √(sin(x)/cos(x)).
The steps for solving this problem are as follows:1. Rewrite the function using trigonometric identities.2. Simplify the function by factoring out a square root and using the Pythagorean identity sin²(x) + cos²(x) = 1.3. Use the substitution method, setting u = cos(x) and du = -sin(x)dx.4. Substitute u back into the original function and integrate.5. Simplify the final answer and include the constant of integration.
The final answer to this calculus problem is √tan(x) = 2√(sin(x)cos(x)) + C, where C is the constant of integration. This can also be written as 2√(sin(2x))/2 + C.
Yes, you can check your solution by differentiating it to see if it equals the original function. In this case, the derivative of 2√(sin(x)cos(x)) is √tan(x). You can also use a graphing calculator to graph both the original function and the antiderivative to see if they match.