# Thread: [SOLVED] hm how does this integral work?

1. ## [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.

2. 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|

3. 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

--Chris

4. 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.

5. ok, i see what u did there, but how the heck did u know to use cscx+cotx for 1!!?!?

6. 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

--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.

7. 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

8. 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.