Question

I need a HERO! INTEGRAL SOS!!!!!

Answer 1

\[\int\limits \frac{ 1 }{ \sin2x }\]

Answer 2

@shifuyanli

Answer 3

@purple_pink

Answer 4

@IrishBoy123 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

Answer 5

@abb0t

Answer 6

sorry i d not know this

Answer 7

hanks though!

Answer 8

ur welcom sorry

Answer 9

@mathstudent55

Answer 10

@dan815

Answer 11

@paki?

Answer 12

will you be my sunshine?

Answer 13

@clynnew

Answer 14

what grade and type of math is this?

Answer 15

nah it's crazy what's shown there

Answer 16

that's horrible

try this
1/sin = sin/sin^2 = sin / (1-cos^2)

sub u = cos2x, du = -2 sin2x dx, dx = -du/2 sin2x
makes it

-∫sin2x/(1-u^2) * du/sin2x
= very do-able! but check algebra , am on hoof at mo, but the idea is good.....

Answer 17

you asked for my help, just showing you one way to do it.

Answer 18

that's calculus for biotechnology BSC

Answer 19

Here's another not-so-obvious approach.
\[\frac{1}{\sin2x}=\csc2x=\csc2x\times\frac{\csc 2x+\cot2x}{\csc2x+\cot2x}=\frac{\csc^2x+\csc2x\cot2x}{\csc2x+\cot2x}\]
Now setting \(u=\csc2x+\cot2x\) gives \(du=(-2\csc2x\cot2x-2\csc^22x)\,dx\), so
\[\int\frac{dx}{\sin2x}=-\frac{1}{2}\int\frac{du}{u}\]

Answer 20

@SithsAndGiggles is there a more obvious way to do that? I'm just a biotech student, our math can't be that hard ehe?

Answer 21

@clynnew thanks! I really appreciate it!

Answer 22

Refer to @IrishBoy123 's post. It's much less time consuming.

Answer 23

but the algebra doesn't make sense @IrishBoy123

Answer 24

dt will give something complicated

Answer 25

\[\frac{1}{\sin2x}=\frac{\sin2x}{\sin^22x}=\frac{\sin2x}{1-\cos^22x}\]
Nothing fancy here, just a trig identity. If \(u=\cos2x\), then \(du=-2\sin2x\,dx\), so
\[\int\frac{dx}{\sin2x}=-\frac{1}{2}\int\frac{du}{1-u^2}\]
You can use partial fractions from here.
\[\frac{1}{1-u^2}=\frac{1}{2}\left(\frac{1}{1-u}+\frac{1}{1+u}\right)\]
\[\int\frac{dx}{\sin2x}=-\frac{1}{4}\int\left(\frac{1}{1-u}+\frac{1}{1+u}\right)\,du\]

Answer 26

thanks a lot!

Answer 27

but why 0.25?

Answer 28

why not 0.5 outside the integral?

Answer 29

@SithsAndGiggles

Answer 30

@clynnew

i must apologise to you.

i think the method you found is horrible, but i certainly recognise and appreciate your kind intentions. mea culpa.