- #1
julian92
- 24
- 0
a bit tricky integral!
Integral of (sqrt(cot[x])dx)
I just need a hint :)
Homework Statement
Integral of (sqrt(cot[x])dx)
Homework Equations
I just need a hint :)
Mark44 said:I believe that an ordinary substitution will work: u = cot(x). At least, that's where I would start.
An identity that might come in handy is cot2(x) + 1 = csc2(x).
physicsman2 said:what I would do is convert cot x to cos x/sin x. Let u = sin x and go from there.
The Chaz said:So... √(du/u) ?
Mark44 said:I think that I would go with a trig substitution next, tan w = u/1.
I haven't worked this through, so caveat emptor!
Cyosis said:This is quite a cumbersome integral. Start out with the substitution given by Mark in post #2 then make a substitution [itex]y=\sqrt{u}[/itex]. Enjoy the partial fractions.
Mark44 said:Here's another tack: Let u = sqrt(cot(x)), du = -csc^2(x)dx/2sqrt(cot(x))
This turns the original integral into -2 int{u^2/(u^4 + 1) * du}
I think that one is amenable to a trig substitution, tan w = u^2
This is a quite messy integral!
Cyosis said:At this step you need to integrate a rational function. To do that use partial fractions.
Cyosis said:You need to use partial fractions to continue from:
[tex]
-2\int \frac{y^2}{y^4+1}
[/tex]
Dickfore said:First, make a substitution:
[tex]
u = \sqrt{\cot(x)}.
[/tex]
Then you will get an integral of a rational function. The denominator is [itex]u^{4} + 1[/itex]. In order to factorize it, you will need all the complex fourth roots of [itex]-1[/itex]. These are:
[tex]
\begin{array}{rcl}
e^{\iota \frac{\pi}{4}} & = & \frac{1 + \iota}{\sqrt{2}} \\
e^{\iota \frac{3 \pi}{4}} & = & \frac{-1 + \iota}{\sqrt{2}} \\
e^{\iotai \frac{5 \pi}{4}} & = & \frac{-1 - \iota}{\sqrt{2}} \\
e^{\iota \frac{7 \pi}{4}} & = & \frac{1 - \iota}{\sqrt{2}}
\end{array}
[/tex]
i.e. you have 2 pairs of complex conjugate roots. Therefore:
[tex]
u^{4} + 1 = (u^{2} + \sqrt{2}u + 1)(u^{2} - \sqrt{2}u + 1)
[/tex]
From here, you should be able to proceed in the standard fashion.
An integral is a mathematical concept that represents the area under a curve on a graph. It can also be thought of as the inverse operation of differentiation.
The "A bit tricky integral!" is difficult because it involves a combination of functions that do not have simple antiderivatives. This makes it challenging to solve using traditional integration techniques.
Yes, there is a specific method for solving the "A bit tricky integral!" called the method of substitution. This involves substituting a variable in the integral with a new variable, making the integral easier to solve.
Yes, the "A bit tricky integral!" can be solved using software or calculators that have integration capabilities. However, it is important to understand the steps involved in solving the integral to ensure accurate results.
The "A bit tricky integral!" is important in science because it is used to solve many real-world problems. In fields such as physics and engineering, integrals are used to calculate important quantities such as work, force, and energy. They are also essential in understanding and modeling natural phenomena.