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

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

 

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.

 

I got an idea! Try

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

 

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.)​

 

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 !

 

@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 ##

 

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

 

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.

 

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.

 

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.

 

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