[SOLVED] hm how does this integral work?

• Oct 9th 2008, 12:24 PM
Legendsn3verdie
[SOLVED] hm how does this integral work?
this one seems to be tricky? can some1 figure out what the book did between these two steps?>?

{integral} 2 cscx dx

= 2 ln( cscx + cotx)

can anybody see what they did?> is it the clever use of 1? i m not exactly sure what went on there. any help appreciated.
• Oct 9th 2008, 12:26 PM
Chop Suey
Quote:

Originally Posted by Legendsn3verdie
this one seems to be tricky? can some1 figure out what the book did between these two steps?>?

{integral} 2 cscx dx

= 2 ln( cscx + cotx)

can anybody see what they did?> is it the clever use of 1? i m not exactly sure what went on there. any help appreciated.

Yes, it is a clever use of 1. Multiply csc(x) by $\frac{\csc{x}+\cot{x}}{\csc{x}+\cot{x}}$

And I believe the answer is -2 ln|cscx + cotx|
• Oct 9th 2008, 12:28 PM
Chris L T521
Quote:

Originally Posted by Legendsn3verdie
this one seems to be tricky? can some1 figure out what the book did between these two steps?>?

{integral} 2 cscx dx

= 2 ln( cscx + cotx)

can anybody see what they did?> is it the clever use of 1? i m not exactly sure what went on there. any help appreciated.

Yes, it is a clever use of 1:

$2\int\csc(x)\,dx=2\int\csc(x)\cdot\frac{\csc(x)+\c ot(x)}{\csc(x)+\cot(x)}\,dx$

Simplify the integrand, and then apply the substitution $u=\csc(x)+\cot(x)$

Take it from here :D

--Chris
• Oct 9th 2008, 12:29 PM
ILoveMaths07
No, it's nothing really.

$\int {\csc x}\ dx = -\ln |\csc x + \cot x|$

So:

$\int {2 \csc x}\ dx$

$= 2 \int {\csc x}\ dx$

$= 2 \cdot -\ln |\csc x + \cot x|$

$= -2 \ln |\csc x + \cot x|.$

I hope that helps. :)

me07.
• Oct 9th 2008, 12:42 PM
Legendsn3verdie
ok, i see what u did there, but how the heck did u know to use cscx+cotx for 1!!?!?
• Oct 9th 2008, 01:06 PM
Legendsn3verdie
Quote:

Originally Posted by Chris L T521
Yes, it is a clever use of 1:

$2\int\csc(x)\,dx=2\int\csc(x)\cdot\frac{\csc(x)+\c ot(x)}{\csc(x)+\cot(x)}\,dx$

Simplify the integrand, and then apply the substitution $u=\csc(x)+\cot(x)$

Take it from here :D

--Chris

ok so u = cscx + cotx

du = -cscx cotx - csc^2 x
= -cscx cotx - (1 + cot^2 x)
= -cscx cotx -1 - cot^2 x

i m not getting the numerator for that substitution..

edit nvm i had it right just went to far, i got it nvm thread solved ty.
• Oct 9th 2008, 01:11 PM
Chris L T521
Quote:

Originally Posted by Legendsn3verdie
ok so u = cscx + cotx

du = -cscx cotx - csc^2 x
= -cscx cotx - (1 + cot^2 x)
= -cscx cotx -1 - cot^2 x

i m not getting the numerator for that substitution..

Before you integrate, you have $\int\frac{{\color{red}\csc^2x+\csc x\cot x}}{\csc x+\cot x}\,dx$

You found du to be $\,du=\left(-\csc^2 x-\csc x\cot x\right) \,dx=-\left({\color{red}\csc^2 x+\csc x\cot x}\right)\,dx$

What integral should you get once you make this substitution?

--Chris
• Oct 9th 2008, 01:20 PM
Legendsn3verdie
Quote:

Originally Posted by Chris L T521
Before you integrate, you have $\int\frac{{\color{red}\csc^2x+\csc x\cot x}}{\csc x+\cot x}\,dx$

You found du to be $\,du=\left(-\csc^2 x-\csc x\cot x\right) \,dx=-\left({\color{red}\csc^2 x+\csc x\cot x}\right)\,dx$

What integral should you get once you make this substitution?

--Chris

yah hm i had it just took it to far, ty.