- #1
chwala
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Homework Statement
[/B]##∫ (1/sin 2x)dx##
Homework Equations
The Attempt at a Solution
let ##u = sin 2x, ⇒ du= 2cos2x dx##
or
##sin 2x= 2 sin x cos x##...[/B]
or
##∫ (1/sin 2x)dx = ∫( csc 2x)dx##
chwala said:Homework Statement
[/B]
##∫ (1/sin 2x)dx##Homework Equations
The Attempt at a Solution
let ##u = sin 2x, ⇒ du= 2cos2x dx##
or
##sin 2x= 2 sin x cos x##...[/B]
or
##∫ (1/sin 2x)dx = ∫( csc 2x)dx##
I would suppose that Math_QED was suggesting that you use u substitution with u = 2x . That does result in ##\ \frac12 \int \csc(u)\ du\ .##chwala said:i don't get how is ## 1/sin 2x = sin x/sin 2x##? and what do you mean by saying substitute 2x?substitute where?
Your question is not clear.chwala said:i have seen it...## 1/2∫cos u du ##= ( -1/2) ln cot x ## why is textbook answer saying tan x?
I assume you use S for ∫ .Cozma Alex said:S (sin^2 x + cos^2 x)/ sin(2x) dx= 1/2 Stanx dx + 1/2 Scotx dx
Then just integrate tangent and co tangent
Yes. I agree!LCKurtz said:@Cozma Alex 's answer gives another way to do ## \int \csc x~dx## that I have never seen before. The usual way is to multiply the integrand by ##\frac{\csc x + \cot x}{\csc x + \cot x}## and let ##u = \csc x + \cot x## to get a ##-\frac {du} u## form. I have never liked this method because if you can remember what to multiply by, you might as well just remember the answer. So now, thinking of Cozma's method, if you need to evaluate ## \int \csc x~dx## just let ##x = 2u## to get$$
\int \csc x~dx = 2\int \csc 2u~du = 2 \int \frac 1 {\sin 2u}~du$$Then use Cozma's method in post #8. Admittedly it is a bit of work, but I like it :-).
chwala said:Thanks Midget any particular reason?
By the way, since Alex's method works so nicely, I didn't follow up on this.SammyS said:Simply use the double angle formula for sin(2x) to give 2sin(x)cos(x) .
You get ##\displaystyle \ \frac12 \int \frac1{\sin(x)\cos(x)}\ dx\ .\ ## Then multiply by either ##\displaystyle\ \frac{\sin(x)}{\sin(x)} \ ## or ##\displaystyle\ \frac{\cos(x)}{\cos(x)} \ .\ ##
Either will work.
It is? I've always done it by multiplying top and bottom by sin(x), converting sin(x)dx to -d cos(x) and writing the denominator as 1-cos2(x). The denominator can then be split using partial fractions to produce two integrals with solutions like ln(1+/-cos(x)).LCKurtz said:The usual way is to multiply the integrand by ##\frac{\csc x + \cot x}{\csc x + \cot x} ##
haruspex said:It is? I've always done it by multiplying top and bottom by sin(x), converting sin(x)dx to -d cos(x) and writing the denominator as 1-cos2(x). The denominator can then be split using partial fractions to produce two integrals with solutions like ln(1+/-cos(x)).
The purpose of integrating ##1/\sin 2x## is to find the area under the curve represented by the function, which can provide valuable information about the behavior and properties of the function.
To integrate ##1/\sin 2x##, you can use the substitution method, where you replace ##2x## with ##u## and rewrite the function in terms of ##u##. Then, you can use the trigonometric identity ##\sin^2 x + \cos^2 x = 1## to simplify the integral and solve for the antiderivative.
The limits of integration for ##1/\sin 2x## depend on the specific problem you are trying to solve. In general, the limits should correspond to the interval over which you want to find the area under the curve represented by the function.
No, you cannot use integration by parts to integrate ##1/\sin 2x## because the function does not have a polynomial term that can be differentiated.
The integration of ##1/\sin 2x## has many real-life applications, such as calculating the average speed of a moving object, finding the period of a pendulum, and determining the amount of work done by a variable force. It is also used in fields such as physics, engineering, and economics to analyze and model various phenomena.