simple integration

[logistic_guy]

here is the question

Solve the integral \(\displaystyle \int \frac{1}{\sin x + 1} dx\).


my attemb
\(\displaystyle u = \sin x\)
\(\displaystyle du = \cos x dx\)
\(\displaystyle \int \frac{1}{\sin x + 1} dx = \int \frac{1}{u + 1}\frac{du}{\cos x}\)

\(\displaystyle \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - u^2 x}\)

\(\displaystyle \int \frac{1}{u + 1}\frac{du}{\sqrt{1 - u^2}}\)
i don't know how to solve this integration

 

[Dr.Peterson]

Solve the integral \(\displaystyle \int \frac{1}{\sin x + 1} dx\).

i don't know how to solve this integration

What else have you tried? Don't just try one thing and give up (especially when you apparently have some reason to call it simple).

But there is, in fact, a very simple way to solve this -- the first thing I tried after skipping what didn't look promising.

Try rewriting the integrand using an identity or two. The idea of multiplying by a sort of conjugate to simplify the denominator might occur to you.

 

[Steven G]

As Dr Peterson hinted, think conjugates.

 

[khansaheb]

here is the question

Solve the integral \(\displaystyle \int \frac{1}{\sin x + 1} dx\).


my attemb
\(\displaystyle u = \sin x\)
\(\displaystyle du = \cos x dx\)
\(\displaystyle \int \frac{1}{\sin x + 1} dx = \int \frac{1}{u + 1}\frac{du}{\cos x}\)

\(\displaystyle \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - u^2 x}\)

\(\displaystyle \int \frac{1}{u + 1}\frac{du}{\sqrt{1 - u^2}}\)
i don't know how to solve this integration

Do you know the process of using "conjugate/s" to simplify algebraic expressions?

 

[khansaheb]

Do you know the process of using "conjugate/s" to simplify algebraic expressions?

Another hint:

conjugate of [sin(x) + 1] is [sin(x)-1]

 

[fresh_42]

Hint: "conjugates" usually refer to the pairing of complex numbers, [imath] a+ i b [/imath] and [imath] a- i b. [/imath] This, and in a more general sense, "conjugates" make use of the formula [imath] (a+b)\cdot (a-b)=a^2-b^2 [/imath] like [imath] (a+i b)\cdot (a-i b)=a^2 - i^2b^2=a^2+b^2. [/imath]

Whenever you see one of the three terms [imath] a+b\, , \,a-b\, , \,a^2 - b^2 [/imath] you should automatically think of this formula, in particular if they show up in an unpleasant place like denominators.

 

[Dr.Peterson]

Another hint:

conjugate of [sin(x) + 1] is [sin(x)-1]

... or [imath]1-\sin(x)[/imath], which I tend to prefer because of what it leads to.

We also use conjugates to "rationalize the denominator" when we see, say, [imath]1+\sqrt{x}[/imath] or [imath]\sqrt{3}-\sqrt{2}[/imath].

 

[fresh_42]

... or [imath]1-\sin(x)[/imath], which I tend to prefer because of what it leads to.

We also use conjugates to "rationalize the denominator" when we see, say, [imath]1+\sqrt{x}[/imath] or [imath]\sqrt{3}-\sqrt{2}[/imath].

Do we require a group of order two when we say conjugates, or does this term also apply if e.g. [imath] \iota^3=1 [/imath] and the automorphisms are of a higher order than two? May we use this term for all inner automorphisms?

 

[logistic_guy]

thank Dr.

What else have you tried? Don't just try one thing and give up (especially when you apparently have some reason to call it simple).

i don't never give up. that's the reason i'm very good in algebra now and soon i'll master calculus too
i call it simple because it contain only one trigonmetric function
also the structure of integration suggest it's very simple
it's just me strugle in basics of integration

But there is, in fact, a very simple way to solve this -- the first thing I tried after skipping what didn't look promising.

what's that? what do you think not promising and you skip?

Try rewriting the integrand using an identity or two. The idea of multiplying by a sort of conjugate to simplify the denominator might occur to you.

i don't see anyhelpful identiy and the term conjugate is reserve to algebra
used in calculus too?
even so it can't be use here without square root
that's my skill from algebra

As Dr Peterson hinted, think conjugates.

thank steven

where is the square root? i don't see it

Do you know the process of using "conjugate/s" to simplify algebraic expressions?

thank khan
no square root in the question

Another hint:

conjugate of [sin(x) + 1] is [sin(x)-1]

i thought conjugate mean something like \(\displaystyle \frac{1}{\sqrt{x + 1}}\) then \(\displaystyle \frac{1}{\sqrt{x + 1}}\times\frac{\sqrt{x - 1}}{\sqrt{x - 1}}\)
the square root condition must be satisfiy but i see you do it anyway without square root. why?

Hint: "conjugates" usually refer to the pairing of complex numbers, [imath] a+ i b [/imath] and [imath] a- i b. [/imath] This, and in a more general sense, "conjugates" make use of the formula [imath] (a+b)\cdot (a-b)=a^2-b^2 [/imath] like [imath] (a+i b)\cdot (a-i b)=a^2 - i^2b^2=a^2+b^2. [/imath]

Whenever you see one of the three terms [imath] a+b\, , \,a-b\, , \,a^2 - b^2 [/imath] you should automatically think of this formula, in particular if they show up in an unpleasant place like denominators.

thank fresh_42
i'm still so confusedas all of you talk about conjugate
algebra tell me to do conjugate when there's square root

 

[fresh_42]

thank fresh_42
i'm still so confusedas all of you talk about conjugate
algebra tell me to do conjugate when there's square root

Just think about [imath] (a+b)\cdot (a-b)=a^2-b^2 [/imath] and in this case, about [imath] (1+\sin)\cdot (1-\sin)=1^2-\sin^2 [/imath] with the identity [imath] 1=\sin^2+\cos^2 [/imath] in mind.

Thinking of the formula [imath] (a+b)\cdot (a-b)=a^2-b^2 [/imath] is an automatism - or at least should be. Four different people in this - so far quite small - thread all have this reflex! The word "conjugate" stems from a deeper algebraic context, namely Galois theory. However, it's not necessary to understand this context. Only that [imath] (a+b)\cdot (a-b)=a^2-b^2 [/imath] should be a reflex is important. "conjugate" is only mathematical slang to abbreviate it.

 

[Dr.Peterson]

what's that? what do you think not promising and you skip?

What was not promising (that is, I could tell it would probably not be helpful) is any substitution like what you tried.

So you try something else. That's how this works.

i don't see anyhelpful identiy and the term conjugate is reserve to algebra

No. Everyone is telling you the opposite. Pay attention.

i thought conjugate mean something like \(\displaystyle \frac{1}{\sqrt{x + 1}}\) then \(\displaystyle \frac{1}{\sqrt{x + 1}}\times\frac{\sqrt{x - 1}}{\sqrt{x - 1}}\)
the square root condition must be satisfiy but i see you do it anyway without square root. why?

No, that particular example would not be helpful anywhere I can think of.

i'm still so confusedas all of you talk about conjugate
algebra tell me to do conjugate when there's square root

This is why I said

The idea of multiplying by a sort of conjugate to simplify the denominator might occur to you.

You only know about the radical conjugate, or perhaps the complex conjugate. But it is a general concept, and you've been told much more specifically how to apply it here. So do it!

Search for the word "conjugate" here:

6.3: Verifying Trigonometric Identities

In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.

math.libretexts.org

 

[fresh_42]

Another possible solution method is the Weierstrass or tangent half-angle substitution [imath] t=\tan \dfrac{x}{2} .[/imath] It is a technique to transform trig functions into polynomials that are often easier to integrate. Here we get (if I made no mistake, it's late here) [math] \dfrac{dx}{1+\sin(x)}=\dfrac{\dfrac{2\,dt}{1+t^2}}{1+\dfrac{2t}{1+t^2}}= 2\cdot \dfrac{dt}{(1+t)^2} [/math]Now you can substitute [imath] u=1+t, [/imath] integrate, and re-substitute [imath] u [/imath] and [imath] t. [/imath]

 

[logistic_guy]

thank Dr. for explaining what you mean
i'll try to use conjugate
\(\displaystyle \frac{1}{\sin x + 1}\frac{\sin x - 1}{\sin x - 1}\)

the denomator have the form fresh_42 give in post \(\displaystyle 10\)
\(\displaystyle \frac{1}{\sin x + 1}\frac{\sin x - 1}{\sin x - 1} = \frac{\sin x - 1}{\sin^2 x - 1^2} = \frac{\sin x - 1}{-\cos^2 x}\)

\(\displaystyle = \frac{1 - \sin x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} = \sec^2 x - \frac{\sin x}{\cos^2 x} \)

\(\displaystyle \int \frac{1}{\sin x + 1} dx = \int \sec^2 x - \frac{\sin x}{\cos^2 x} dx = \tan x - \int \frac{\sin x}{\cos^2 x} dx\)

\(\displaystyle u = \cos x\)
\(\displaystyle du = -\sin x dx\)

\(\displaystyle -\int \frac{\sin x}{\cos^2 x} dx = \int \frac{1}{u^2} du = -\frac{1}{u} + c = -\frac{1}{\cos x} + c = -\sec x + c\)

this mean \(\displaystyle \int \frac{1}{\sin x + 1} dx = \tan x - \sec x + c\)

Another possible solution method is the Weierstrass or tangent half-angle substitution [imath] t=\tan \dfrac{x}{2} .[/imath] It is a technique to transform trig functions into polynomials that are often easier to integrate. Here we get (if I made no mistake, it's late here) [math] \dfrac{dx}{1+\sin(x)}=\dfrac{\dfrac{2\,dt}{1+t^2}}{1+\dfrac{2t}{1+t^2}}= 2\cdot \dfrac{dt}{(1+t)^2} [/math]Now you can substitute [imath] u=1+t, [/imath] integrate, and re-substitute [imath] u [/imath] and [imath] t. [/imath]

let me try this idea

this mean \(\displaystyle \int \frac{1}{\sin x + 1} dx = \int \dfrac{\dfrac{2}{1+t^2}}{1+\dfrac{2t}{1+t^2}} dt = \int 2\cdot \dfrac{1}{(1+t)^2} dt\)

\(\displaystyle u = 1 + t\)
\(\displaystyle du = dt\)

\(\displaystyle \int 2\cdot \dfrac{1}{(1+t)^2} dt = 2\int \frac{1}{u^2}du = -2\frac{1}{u} + c = -2\frac{1}{1 + t} + c = \frac{-2}{1 + \tan\frac{x}{2}} + c\)

i get two different answers
i check my calculations \(\displaystyle 3\) timesi can't see where my mistake

 

[fresh_42]

I cannot find a mistake. Let's double-check it.

WA confirms the Weierstrass substitution method since
[math] \displaystyle{\dfrac{2}{1+\tan(x/2)}} + \displaystyle{\int \dfrac{dx}{1+\sin(x)}} =2 [/math]which vanishes in the constant. Now to the other method. We get again from WA
[math] \tan(x)-\sec(x) - \displaystyle{\int \dfrac{dx}{1+\sin(x)}} =-1 [/math]which also vanishes in the constant.

Sorry for being lazy one more time. WA finally confirms
[math] \tan(2x)-\sec(2x) + \dfrac{2}{1+ \tan(x)} = 1.[/math]I leave it to you to formally prove this identity.

 

[logistic_guy]

I cannot find a mistake. Let's double-check it.

WA confirms the Weierstrass substitution method since
[math] \displaystyle{\dfrac{2}{1+\tan(x/2)}} + \displaystyle{\int \dfrac{dx}{1+\sin(x)}} =2 [/math]which vanishes in the constant. Now to the other method. We get again from WA
[math] \tan(x)-\sec(x) - \displaystyle{\int \dfrac{dx}{1+\sin(x)}} =-1 [/math]which also vanishes in the constant.

Sorry for being lazy one more time. WA finally confirms
[math] \tan(2x)-\sec(2x) + \dfrac{2}{1+ \tan(x)} = 1.[/math]I leave it to you to formally prove this identity.

i'm not understand what you're trying to say here

do you say
1. my first solution wrong and my second solution right
2. my first solution right and my second solution wrong
3. both are wrong

 

[Dr.Peterson]

thank Dr. for explaining what you mean
i'll try to use conjugate
\(\displaystyle \frac{1}{\sin x + 1}\frac{\sin x - 1}{\sin x - 1}\)

the denomator have the form fresh_42 give in post \(\displaystyle 10\)
\(\displaystyle \frac{1}{\sin x + 1}\frac{\sin x - 1}{\sin x - 1} = \frac{\sin x - 1}{\sin^2 x - 1^2} = \frac{\sin x - 1}{-\cos^2 x}\)

\(\displaystyle = \frac{1 - \sin x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} = \sec^2 x - \frac{\sin x}{\cos^2 x} \)

\(\displaystyle \int \frac{1}{\sin x + 1} dx = \int \sec^2 x - \frac{\sin x}{\cos^2 x} dx = \tan x - \int \frac{\sin x}{\cos^2 x} dx\)

\(\displaystyle u = \cos x\)
\(\displaystyle du = -\sin x dx\)

\(\displaystyle -\int \frac{\sin x}{\cos^2 x} dx = \int \frac{1}{u^2} du = -\frac{1}{u} + c = -\frac{1}{\cos x} + c = -\sec x + c\)

this mean \(\displaystyle \int \frac{1}{\sin x + 1} dx = \tan x - \sec x + c\)

Or you could have recognized the second part earlier:

\(\displaystyle \frac{1}{1+\sin x}=\frac{1}{1+\sin x}\frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{1-\sin^2 x} = \frac{1-\sin x}{\cos^2 x}= \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} = \sec^2 x-\sec x\tan x\)

which can be integrated immediately.

i get two different answers

Never say this too quickly when you are integrating; you can get two answers that look very different, but are equivalent. I always expect that.

Both of your answers are CORRECT!

 

[logistic_guy]

Both of your answers are CORRECT!

how?

they're very different

 

[fresh_42]

how?

they're very different

You got [imath] \displaystyle{\int \dfrac{dx}{1+\sin(x)}}=f(x)+C =g(x)+C' [/imath] and [imath] f(x)-g(x)=C''=C'-C [/imath] so where is the problem?

[imath] \sec(x)=\dfrac{1}{\cos(x)} [/imath] or [imath] \cos(2x)=1-\sin^2(x) [/imath] also look very different. Nevertheless, the equations are correct. Particularly trig functions can often be expressed in more than one way.

 

[logistic_guy]

You got [imath] \displaystyle{\int \dfrac{dx}{1+\sin(x)}}=f(x)+C =g(x)+C' [/imath] and [imath] f(x)-g(x)=C''=C'-C [/imath] so where is the problem?

[imath] \sec(x)=\dfrac{1}{\cos(x)} [/imath] or [imath] \cos(2x)=1-\sin^2(x) [/imath] also look very different. Nevertheless, the equations are correct. Particularly trig functions can often be expressed in more than one way.

after many trials i discover
\(\displaystyle \tan x - \sec x = \frac{-2}{1 + \tan \frac{x}{2}} + 1\)

if i write \(\displaystyle \frac{-2}{1 + \tan \frac{x}{2}} + 1\) in terms of \(\displaystyle \sin x\) and \(\displaystyle \cos x\)
i still don't get the expression \(\displaystyle \tan x - \sec x\)

 

[fresh_42]

if i write \(\displaystyle \frac{-2}{1 + \tan \frac{x}{2}} + 1\) in terms of \(\displaystyle \sin x\) and \(\displaystyle \cos x\)
i still don't get the expression \(\displaystyle \tan x - \sec x\)

I use the half-angle formulas from

Formelsammlung Trigonometrie – Wikipedia

de.wikipedia.org

[math]\begin{array}{lll} 1-\dfrac{2}{1+\tan(x/2)}&=1-\dfrac{2}{1+\dfrac{\sin(x/2)}{\cos(x/2)}}\\[30pt] &=1-\dfrac{2\cos(x/2)}{\sin(x/2)+\cos(x/2)}\\[24pt] &=\dfrac{\sin(x/2)+\cos(x/2)-2\cos(x/2)}{\sin(x/2)+\cos(x/2)}\\[24pt] &=\dfrac{\sin(x/2)-\cos(x/2)}{\sin(x/2)+\cos(x/2)}\\[24pt] &=\dfrac{\sqrt{\dfrac{1-\cos(x)}{2}}-\sqrt{\dfrac{1+\cos(x)}{2}}}{\sqrt{\dfrac{1-\cos(x)}{2}}+\sqrt{\dfrac{1+\cos(x)}{2}}}\\[24pt] &=\dfrac{\sqrt{1-\cos(x)}-\sqrt{1+\cos(x)}}{\sqrt{1-\cos(x)}+\sqrt{1+\cos(x)}}\cdot \dfrac{\sqrt{1-\cos(x)}-\sqrt{1+\cos(x)}}{\sqrt{1-\cos(x)}-\sqrt{1+\cos(x)}} \\[24pt] &=\dfrac{\left(\sqrt{1-\cos(x)}-\sqrt{1+\cos(x)}\right)^2}{ \left( \sqrt{1-\cos(x)} \right)^2 - \left( \sqrt{1+\cos(x)} \right)^2 }\\[24pt] &=\dfrac{(1-\cos(x))-2\sqrt{(1-\cos(x))(1+\cos(x))}+(1+\cos(x))}{-2\cos(x)}\\[24pt] &=-\dfrac{1}{\cos(x)}+\dfrac{\sqrt{1-\cos^2(x)}}{\cos(x)}\\[24pt] &=-\sec(x) + \dfrac{\sqrt{\sin^2(x)}}{\cos(x)}\\[24pt] &=\tan(x)-\sec(x) \end{array}[/math]
There might be a shorter way with other formulas but I was happy to find one. Please study the techniques I used from line to line. You should be able to do such calculations on your own!