# Integration problem in trigonometric functions

1. Nov 25, 2013

### kashan123999

1. The problem statement, all variables and given/known data

∫(1/sinx+ cosx) dx

2. Relevant equations

i dont know any proper substitution in this question,maybe there are none

3. The attempt at a solution

2. Nov 25, 2013

### brmath

better to integrate the terms separately. The cos(x) is no problem. The other term is csc(x). Here's how you go about it:

$\int$ csc(x) dx =$\int$ csc(x) $\cdot \frac{csc x + cot x}{ csc x + cot x}$dx
use the substitution:

u = csc x + cot x

You say that's not obvious? Right -- it is a trick. There are thousands of them, worked out over many years by lots of smart mathematicians. This one is more commonplace than some.

3. Nov 25, 2013

### Staff: Mentor

brmath, I don't see that this will help.
I believe that this is the integral in question:
$$\int \frac{dx}{sin(x) + cos(x)}$$

kashan, brmath interpreted what you wrote (1/sin(x)) + cos(x), which from your work, wasn't what you intended. To prevent this kind of confusion, place parentheses around the entire denominator, like this: 1/(sin(x) + cos(x))

Last edited: Nov 25, 2013
4. Nov 25, 2013

### R136a1

5. Nov 25, 2013

### brmath

I did so interpret it. Sorry.

6. Nov 25, 2013

### LCKurtz

No need to apologize. You worked what was actually posted. kashan123999 caused the confusion with his improperly written post.

7. Nov 25, 2013

### Staff: Mentor

Yes, no need for apologies, brmath. It takes some practice here to understand what neophytes mean, as opposed to what they actually write. The only way I could figure out what the OP actually meant was from the work he showed.

8. Nov 25, 2013

### lurflurf

hint
$$\frac{1}{\sin(x)+\cos(x)}=\frac{\sqrt{2}}{2}\csc (x+\pi/4)$$

9. Nov 25, 2013

### Staff: Mentor

As one of my calculus instructors was fond of saying, "Intuitively obvious to the most casual observer."

10. Nov 25, 2013

### iRaid

Well then, this makes it quite simple doesn't it...

11. Nov 25, 2013

### brmath

His idea of casual observer being not exactly applicable to people like me.

12. Nov 25, 2013

### LCKurtz

Has anyone but me noticed the OP never returned to this thread?

13. Nov 26, 2013

### Staff: Mentor

So let's hold with any more help until we hear back from the OP.
I assure you, he was speaking ironically...

14. Nov 26, 2013

### brmath

of course he was speaking ironically -- so was I.

15. Dec 1, 2013

### kashan123999

I am extremely sorry for that misplacement of parenthesis that caused the whole confusion,very very sorry to all of you...it is yes it is actually ∫[1/(sinx + cosx)] dx so then,i have corrected it,can anyone tell me some easy or intuitive substitution,cause our teacher substituted 1 for some angle of sin i guess...and i am not able to grasp that...

16. Dec 1, 2013

### LCKurtz

Look at post #4 and post #8. Two different ways to attack the problem. You might have additional questions after looking at those posts.

17. Dec 1, 2013

### lurflurf

(sinx-cosx)/[(sinx)^2 - (cosx^2)]
=sinx/[(sinx)^2 - (cosx^2)]-cosx/[(sinx)^2 - (cosx^2)]
you can substitute u=sin(x)in one and u=cos(x) in the other and use
1=sin^2(x)+cos^2(x)
and
the integration formula for 1/(1-a^2 x^2)
or use my identity above
$$\frac{1}{\sin(x)+\cos(x)}=\frac{\sqrt{2}}{2}\csc (x+\pi/4)$$
along with the integration formula for csc(x)