Integration of ##1/\sin 2x##

[chwala]

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
$∫ (1/sin 2x)dx = ∫( csc 2x)dx$

 

[member 587159]

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
$∫ (1/sin 2x)dx = ∫( csc 2x)dx$

First, substitute the 2x. Then, notice that 1/sinx = sinx/sin^2x.

 

[chwala]

i don't get how is $1/sin 2x = sin x/sin 2x$? and what do you mean by saying substitute 2x?substitute where?

 

[chwala]

i have seen it........$1/2∫cos u du$= ( -1/2) ln cot x ## why is textbook answer saying tan x?

 

[SammyS]

i don't get how is $1/sin 2x = sin x/sin 2x$? and what do you mean by saying substitute 2x?substitute where?

I would suppose that Math_QED was suggesting that you use u substitution with u = 2x . That does result in $\ \frac12 \int \csc(u)\ du\ .$

However, I would make a different suggestion.
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.

 

[physics user1]

I got an idea! Try

1= sin^2 x + cos^2 x and divide the fraction in two parts :)

 

[SammyS]

i have seen it........$1/2∫cos u du$= ( -1/2) ln cot x ## why is textbook answer saying tan x?

Your question is not clear.

Do you mean that Math_QED's suggestion gives you $\ 1/2∫\cos u\ du \ ?\$ If so, then no, that results in $\ 1/(2∫(1/\sin u)\ du) \ .\$

Furthermore, $( -1/2) \ln\, (\cot x)$ is indeed equal to $( 1/2) \ln\, (\tan x)$

(Alex has a rather interesting idea !

Added in Edit:
Alex's idea doesn't work out the way I at first thought, but maybe it's still helpful, maybe not.)​

 

[physics user1]

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

 

[SammyS]

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

I assume you use S for ∫ .

OK!

Yes, that is very cleaver !

 

[LCKurtz]

@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 :-).

 

[SammyS]

@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 :-).

Yes. I agree!

As far as I can see, this doesn't work so nicely for integrating sec(x). The only way I could get that integral to work in a similar fashion was to use the co-function identity along with u-substitution.

$\displaystyle \int \sec x \,dx=\int \csc \left(\frac{\pi}{2}-x\right) \,dx$

 

[chwala]

nice discussion...looks like there are different ways of doing it...i would vouch for the double angle way...

 

[MidgetDwarf]

I would just use the identity $$1/sinx=cscx$$

 

[chwala]

Thanks Midget any particular reason?

 

[MidgetDwarf]

Thanks Midget any particular reason?

the integration of that identity is $$ln|cscx-cotx|$$.
There are different proofs for it you can do. The nicest one is using an identity to rewrite cscx, then replace it in the integral. The longer way, without knowing the trick, is to do parts.

 

[SammyS]

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.

By the way, since Alex's method works so nicely, I didn't follow up on this.

Saying to multiply by either $\displaystyle\ \frac{\sin(x)}{\sin(x)} \$ or $\displaystyle\ \frac{\cos(x)}{\cos(x)} \$ was a bit misleading although technically correct.

Rather multiply by either $\displaystyle\ \frac{\displaystyle 1 }{\ \frac{\sin(x)}{\sin(x)}\ } \$ or $\displaystyle\ \frac{\displaystyle 1 }{\frac{\ \cos(x)}{\cos(x)\ }} \ .\$

For the former this gives: $\displaystyle \ \frac{1}{2} \int \frac1{\sin^2(x)\cot(x)}\ dx\$

$\displaystyle = \frac{1}{2} \int \frac{\csc^2(x)}{\cot(x)}\ dx\$​

This is easily handled with u-substitution.

 

[chwala]

nice one Sammy..

 

[haruspex]

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)).

 

[LCKurtz]

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)).

Heh. By "the usual way" I mean the way I learned it back in the '50's.

 

[Irene Kaminkowa]

Another way
$$
\int \frac{dx}{sin(2x)} = \int \frac{dx}{2sinx~cosx} = \int \frac{dx}{2cos^2 x~tanx} = \begin{bmatrix}
t = tan x\\
dt = \frac{dx}{cos^2x}
\end{bmatrix} = \frac{1}{2}\int \frac{dt}{t}= \frac{1}{2}ln|t| + C = \frac{1}{2}ln|tan x| + C
$$