1. ## Trigo Integrals

kindly check... this is an integral without using the formula of trig integrals:

$\int \cot(5x) + \csc(5x) dx$

solution:
$\int \frac{\cos(5x)}{\sin(5x)} + \csc(5x) dx$
from the first integral
let u = sin(5x)
du = cos(5x)dx

from the right integral u multiply both denominator and numerator by (cot(5x) + csc(5x))
$\frac{1}{5}\ln{(\sin{5x})} + \int \frac{\csc(5x)\cot(5x) + \csc^2(5x)}{\cot(5x)+\csc(5x)} dx$

let u = $\cot(5x)+\csc(5x)$
du = $-5(\csc^2(5x) + \csc(5x)\cot(5x)) dx$

integrated to :
$\frac{1}{5}\ln{(\sin{5x})} - \frac{1}{5}\ln(\cot(5x) + \csc(5x))$

i ended up having
$\frac{1}{5} \ln(\frac{\sin^2(5x)}{\cos(5x) + 1}) + C$

the answer at back of book is supposed to be:
$\frac{1}{5} \ln (1-\cos(5x)) + C$

and you shud not use a formula of trig integral

2. You could use the identity: $cot(5x)-csc(5x)=-tan(\frac{5x}{2})$

3. is that half angled identity?
im confused where the 2 appear

4. Yep, it's a half-angle.

$tan(\frac{x}{2})=\frac{1-cos(x)}{sin(x)}=csc(x)-cot(x)$

Therefore, $cot(x)-csc(x)=-tan(\frac{x}{2})$

5. but the trig formulas are not allowed:
example : integral of tan(x) = |ln sec(x)| + C
the formula of integrals tan, cot and sec and csc are not allowed becuz it is not yet discsed...
is my solution right? i think the only problem i hav is on logarithm & trig identity

6. waiiiitttt

the question is supposed to be $\int{\cot(5x) +\csc(5x)dx}$
but it has still the same solution ive written

7. After looking at your answer, you have it correct. It just needs hammered into the book form. Sometimes the books can simplify beyond all recognition(to paraphrase Soroban).

kindly check... this is an integral without using the formula of trig integrals:

$\int \cot(5x) + \csc(5x) dx$

solution:
$\int \frac{\cos(5x)}{\sin(5x)} + \csc(5x) dx$
from the first integral
let u = sin(5x)
du = cos(5x)dx

from the right integral u multiply both denominator and numerator by (cot(5x) + csc(5x))
$\frac{1}{5}\ln{(\sin{5x})} + \int \frac{\csc(5x)\cot(5x) + \csc^2(5x)}{\cot(5x)+\csc(5x)} dx$

let u = $\cot(5x)+\csc(5x)$
du = $-5(\csc^2(5x) + \csc(5x)\cot(5x)) dx$

integrated to :
$\frac{1}{5}\ln{(\sin{5x})} - \frac{1}{5}\ln(\cot(5x) + \csc(5x))$

i ended up having
$\frac{1}{5} \ln(\frac{\sin^2(5x)}{\cos(5x) + 1}) + C$

the answer at back of book is supposed to be:
$\frac{1}{5} \ln (1-\cos(5x)) + C$

and you shud not use a formula of trig integral
Three things.
1) I can see nothing wrong with your work.

3) As far as I can tell the two answers do not differ by a constant.

I think your book is wrong.

$\frac{1}{5} \ln(\frac{\sin^2(5x)}{|\cos(5x) + 1|}) + C$

But you have the idea.

-Dan

9. When I run it through my TI I get $\frac{1}{5}ln(cos(5x)-1)$

10. Originally Posted by topsquark
Three things.
1) I can see nothing wrong with your work.

3) As far as I can tell the two answers do not differ by a constant.

I think your book is wrong.

$\frac{1}{5} \ln(\frac{\sin^2(5x)}{|\cos(5x) + 1|}) + C$

But you have the idea.

-Dan
Originally Posted by galactus
When I run it through my TI I get $\frac{1}{5}ln(cos(5x)-1)$
Okay I see my mistake now, but I still can't convince my calculator they are the same for some reason.

$\frac{1}{5} \ln(\frac{\sin^2(5x)}{|\cos(5x) + 1|}) = \frac{1}{5} \ln(\frac{1 - \cos^2(5x)}{|\cos(5x) + 1|})$

$= \frac{1}{5} \ln(\frac{(1 - cos(5x))(1 + cos(5x))}{|\cos(5x) + 1|})$

$= \frac{1}{5}ln|1 - cos(5x)| = \frac{1}{5}ln|cos(5x) - 1|$

So the two answers are the same. But why can't I plug an x in my calculator to see that the two expressions are the same??

-Dan

Edit: No problems with the calculator difficulty. I still had my calculator on "exact." The differences between calculating the two expressions are infinitesimal, as expected.

11. We can use well-known results.

12. Originally Posted by galactus
When I run it through my TI I get $\frac{1}{5}ln(cos(5x)-1)$
However cos(5x)-1 is negative, which is not allowed for ln(). So you must take the absolute value or 1-cos(5x) instead.

13. thank u