1. ## Integration

$\int{csc2x}dx$

1. I used the identity $csc2x=\frac{1}{2\sin{x}\cos{x}}$
I tried letting $u=\sin{x}$. So

$\frac{1}{2}\int{\frac{1}{u-u^3}}du$.

I need an explanation from top to bottom guys, please.

I'm purposefully trying not to use partial fraction decomp.

2. $\int csc(2x)dx=\int sec(\frac{\pi}{2}-2x)dx$

Now, integrate the sec. Know that one?.

3. Originally Posted by galactus
Know that one?.
Uhhhh.....

Let's say I was going to employ decomp then. How would I do it?

4. Hello, VonNemo19!

There is a formula for this problemL . $\int\csc\theta\,d\theta \;=\;\ln|\csc\theta - \cot\theta| + C$.
No one has taught it to you?

$\int\csc2x\,dx$
You can derive the formula like this . . .

We have: . $\int \csc\theta\,d\theta$

Multiply by $\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}\!:\quad \int\frac{\csc\theta}{1}\cdot\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}\,d\theta$ . $=\;\int\frac{\csc^2\!\theta + \csc\theta\cot\theta}{\csc\theta + \cot\theta}\,d\theta$

Let $u \:=\:\csc\theta + \cot\theta \quad\Rightarrow\quad du \:=\:(-\csc\theta\cot\theta - \csc^2\!\theta)\,d\theta$

. . Hence: . $(\csc^2\!\theta + \csc\theta\cot\theta)\,d\theta \:=\:-du$

Substitute: . $-\int\frac{du}{u} \;=\;-\ln|u| + C$

Back-substitute: . $-\ln|\csc\theta + \cot\theta| + C \;=\; \ln|\csc\theta + \cot\theta|^{-1} + C \;=\; \ln\left|\frac{1}{\csc\theta + \cot\theta}\right| + C$

Multiply by $\frac{\csc\theta - \cot\theta}{\csc\theta - \cot\theta}\!:\quad \ln\left|\frac{1}{\csc\theta + \cot\theta}\cdot\frac{\csc\theta - \cot\theta}{\csc\theta-\cot\theta}\right| + C$ . $= \;\ln\left|\frac{\csc\theta - \cot\theta}{\csc^2\!\theta - \cot^2\!\theta}\right| + C$

Since $\csc^2\!\theta-\cot^2\!\theta \:=\:1$, we have: . $\ln|\csc\theta - \cot\theta| + C$

Now apply this formula to your problem . . .

5. Multiply by $\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}\!:\quad \int\frac{\csc\theta}{1}\cdot\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}\,d\theta$
This is the approach that I have seen, but can someone explain why it is the most ideal as it seems rather non-intuitive to me.

6. Here is another way. More involved.

$\int csc(2x)dx$

$\int\frac{1}{sin(2x)}dx$

$\int\frac{1}{sin(2x)}\cdot\frac{sin(2x)}{sin(2x)}d x$

$\int\frac{sin^{2}(2x)}{1-cos^{2}(2x)}dx$

Let $u=cos(2x), \;\ \frac{-du}{2}=sin(2x)dx$

This gives:

$\frac{-1}{2}\int\frac{1}{1-u^{2}}du$

$\frac{1}{4}\int\frac{1}{u-1}du-\frac{1}{4}\int\frac{u+1}du$

$\frac{ln(u-1)}{4}-\frac{ln(u+1)}{4}$

$\frac{1}{4}ln(\frac{u-1}{u+1})$

Resub:

$\frac{1}{4}ln(\frac{cos(2x)-1}{cos(2x)+1})$

Inside the parentheses, multiply top and bottom by $cos(2x)+1$

(one can also use $cos(2x)-1$)

This gives:

$\frac{1}{4}ln(\frac{sin^{2}(2x)}{(cos(2x)+1)^{2}})$

Without further detail, the identity inside the parentheses is equal to $tan^{2}(x)$

and we get:

$\frac{1}{4}ln(tan^{2}(x))$

By the property of logs:

$\boxed{\frac{1}{2}ln(tan(x))}$

7. Could I do it this way?

$\int\csc{2x}dx=\int\frac{1}{2\sin{x}\cos{x}}dx$

Let $u=\sin{x}$, then $du=\cos{x}dx$

$\int{\frac{1}{2\sin{x}\cos{x}}}dx=\int\frac{1}{2u\ cos^2{x}}du$ $=\int\frac{1}{2u(1-\sin^2{x})}du=\int\frac{1}{2u(1-u^2)}du=\int\frac{1}{2u(1-u)(1+u)}du$

But, $\frac{1}{2u(1-u)(1+u)}=\frac{A}{2u}+\frac{B}{1-u}+\frac{C}{1+u}$

$1=A(u-1)(u+1)+B2u(u+1)+C2u(1-u)\Rightarrow{A}=-1,B=\frac{1}{4},C=-\frac{1}{4}$

Then $\int\csc2xdx=-\frac{1}{2}\ln{|u|}+\frac{1}{4}\ln{|1-u|}-\frac{1}{4}\ln{|1+u|}+C$

If I sub back for u...

8. Originally Posted by VonNemo19
$1=A(u-1)(u+1)+B2u(u+1)+C2u(1-u)\Rightarrow{A}=-1,B=\frac{1}{4},C=\frac{1}{4}$
Shouldn't that be $1=A(1-u)(1+u)+B2u(1+u)+C2u(1-u)$ which gives $A=1, B=\frac {1}{4}, C=\frac{-1}{4}$? At least that's my thought.

9. Or

$\int {\csc 2xdx} = \int {\frac{{dx}}
{{\sin 2x}}} = \frac{1}
{2}\int {\frac{{dx}}
{{\sin x\cos x}}} = \frac{1}
{2}\int {\frac{{{{\sin }^2}x + {{\cos }^2}x}}
{{\sin x\cos x}}dx} =$

$= \frac{1}
{2}\int {\frac{{\sin x}}
{{\cos x}}} dx + \frac{1}
{2}\int {\frac{{\cos x}}
{{\sin x}}} dx = - \frac{1}
{2}\int {\frac{{d\left( {\cos x} \right)}}
{{\cos x}}} + \frac{1}
{2}\int {\frac{{d\left( {\sin x} \right)}}
{{\sin x}}} =$

$= - \frac{1}
{2}\ln \left| {\cos x} \right| + \frac{1}
{2}\ln \left| {\sin x} \right| + C = \frac{1}
{2}\ln \left| {\tan x} \right| + C.$

10. Can anyone verify my last post?

11. your last post looks good, but I'd highly recommend memorizing the general csc integration formula that Soroban posted

12. $\int \csc(2x) \ dx = \int \frac{dx}{\sin(2x)} = \int \frac{1+\tan^{2}(x)}{2\tan(x)} \ dx$

let $u = \tan(x)$

then $x = \arctan(u)$

and $dx = \frac{du}{1+u^{2}}$

$= \int \frac{1+u^{2}}{2u} \frac{du}{1+u^{2}} = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2}ln(|u|) + C = \frac{1}{2}\ln(|\tan(x)|) + C$

13. Originally Posted by artvandalay11
...but I'd highly recommend memorizing the general csc integration formula that Soroban posted
Why? There clearly is abundance of simpler ones. Including the one I derived. It involves only sine. Is there a good reason why soroban's is the best?

14. Originally Posted by VonNemo19
Why? There clearly is abundance of simpler ones. Including the one I derived. It involves only sine. Is there a good reason why soroban's is the best?
Soroban's identity is elegant, and standard. It's good to know the integration formula for every trigonometric function. More importantly, the solution you derived was for a different integral, and it only (directly) involves the sine function because there was a double angle involved. In other words, he was suggesting you derive your result from the formula for

$\int\csc x\, dx,$

whereas you derived from scatch a formula for

$\int \csc 2x\, dx,$

which is somewhat different.

Incidentally, and apropos of nothing, do you know the method of Weierstrass substitution? It transforms any rational function in terms of sine and cosine into an ordinary rational function in a single variable.

15. Originally Posted by AlephZero
More importantly, the solution you derived was for a different integral, and it only (directly) involves the sine function because there was a double angle involved.
Indeed! I can't believe I missed that. Thank you.

Originally Posted by AlephZero
Incidentally, and apropos of nothing, do you know the method of Weierstrass substitution?
No, but I vaguely recall writing trig functions as algebraic functions; or was that inverse trig functions...?

I can't remember. I do know that the domain would have to be restricted...

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