- #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##
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\ .##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.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 ∫ .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!@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 :-).
Thanks Midget any particular reason?
By the way, since Alex's method works so nicely, I didn't follow up on this.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)).The usual way is to multiply the integrand by ##\frac{\csc x + \cot x}{\csc x + \cot x} ##
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)).