Integration problem in trigonometric functions

[kashan123999]

Homework Statement

please help me with this integration problem?

∫(1/sinx+ cosx) dx

Homework Equations

i don't know any proper substitution in this question,maybe there are none

The Attempt at a Solution

i tried rationalizing and it has got me this far.. ∫(sinx-cosx)/[(sinx)^2 - (cosx^2)] dx...please help me

 

[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.

 

[Mark44]

Homework Statement

please help me with this integration problem?

∫(1/sinx+ cosx) dx

Homework Equations

i don't know any proper substitution in this question,maybe there are none

The Attempt at a Solution

i tried rationalizing and it has got me this far.. ∫(sinx-cosx)/[(sinx)^2 - (cosx^2)] dx...please help me

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.

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))

 

[R136a1]

The world's sneakiest substitution: http://en.wikipedia.org/wiki/Tangent_half-angle_substitution

 

[brmath]

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))

I did so interpret it. Sorry.

 

[LCKurtz]

I did so interpret it. Sorry.

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

 

[Mark44]

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.

 

[lurflurf]

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

 

[Mark44]

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

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

 

[iRaid]

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

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

 

[brmath]

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

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

 

[LCKurtz]

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

 

[Mark44]

So let's hold with any more help until we hear back from the OP.

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

I assure you, he was speaking ironically...

 

[brmath]

So let's hold with any more help until we hear back from the OP.

I assure you, he was speaking ironically...

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

 

[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...

 

[LCKurtz]

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...

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

 

[lurflurf]

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...

From your step
(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)