$\int \frac{dx}{1-\sin x+\cos x}$

Question

Solve :$$ \int \frac{dx}{1-\sin x+\cos x} $$

I tried:

$$\int \frac{\text{dx}}{(1+\cos x)-\sin x} \times \frac{(1+\cos x)+\sin x}{(1+\cos x)+ \sin x} dx=\int \frac{1+\cos x+\sin x}{(1+\cos x)^2-(\sin x)^2}$$ $$=\int \frac{1+\cos x+\sin x}{1-\sin x^2+2 \cos x+\cos x^2}=\int \frac{1+\cos x+\sin x}{2 \cos x(\cos x+1)}$$ $$=\int (\frac{1}{2 \cos x}+ \frac{\sin x}{2 \cos x(\cos x+1)})dx=\frac{1}{2} \ln(\sec x+\tan x)+ \int \frac{\sin x}{2\cos x(\cos x+1)}dx$$ $ \cos x=u$ , $du=-\sin xdx$ $$\int \frac{-du}{2u(u+1)}= \frac{-1}{2} \int (\frac{1}{u}-\frac{1}{u+1})= \frac{-1}{2}(\ln(\cos x)-\ln(\cos x+1))$$

final answer : $$\frac{1}{2} \ln(\sec x+\tan x)-\frac{1}{2} \ln(\frac{\cos x}{1+\cos x})+ c$$

First: Is my answer right?

Second: Is there another approach or easier approach to solve this integral?

"Is my answer right?" Have you tried taking the derivative of the result to check that? — , Oct 20 '19 at 19:56
You can use https://www.integral-calculator.com/ : it gives this answer. — Jean Marie, Oct 20 '19 at 20:04
now I tried to calculating the derivative of the result is long and a little hard! — Aligator, Oct 20 '19 at 20:05
@Jean Marie The integral calculator method to solve this integral was really far from the mind and my approach was a little long! — Aligator, Oct 20 '19 at 20:13
Yes, it uses implicitly formulas $\sin x =\dfrac{2t}{1+t^2}$ etc... with $t=\tan \tfrac{x}{2}$ you may know. — Jean Marie, Oct 20 '19 at 20:17
$$ \int \frac{dx}{1-\sin x+\cos x} = \coth^{-1}\frac{2-\sin x +\cos x}{\sin x + \cos x}$$ — Quanto, Apr 11 '23 at 21:19

score 4 · Answer 1 · answered Oct 20 '19 at 20:13

4

Your answer is correct

You may try the following as well.

Substitute $$\sin x =\frac {2\tan (x/2)}{1+\tan^2(x/2)}$$ and $$\cos x =\frac {1-\tan^2 (x/2)}{1+\tan^2(x/2)}$$

Then let $u=\tan (x/2)$

answered Oct 20 '19 at 20:13

Mohammad Riazi-Kermani

69,891

Thanks, it is more simple and shorter – Aligator Oct 20 '19 at 20:19
Thank you for your attention. – Mohammad Riazi-Kermani Oct 20 '19 at 20:21
The point is: most of the time When we see $sinx$ or $cosx$ or both in denominator using $u=\tan (x/2)$ substitution can be helpful right? – Aligator Oct 20 '19 at 20:22
It worths trying any way.Specially in linear combination of $\sin x$ and $\cos x$ – Mohammad Riazi-Kermani Oct 20 '19 at 20:23

score 3 · Answer 2 · answered Oct 22 '19 at 19:32

3

My hint for your question:

$$\int \frac{dx}{1-\sin x+\cos x}$$ If I apply the $t-$substitution $t=\tan(\frac{x}{2})$ you will have $$\int \frac{dx}{1-\sin x+\cos x}=\int \frac{1}{-u+1}du$$ Using another $y-$substitution for last integral of the type $p=-y+1$ you will have: $$\int \frac{1}{-u+1}du=\int -\frac{1}{p}dp$$ Hence remebering all substitutions $p=-u+1,\,t=\tan (\frac{x}{2} )$ you will obtain $$\int \frac{dx}{1-\sin x+\cos x}=-\ln \left|-\tan \left(\frac{x}{2}\right)+1\right|+k, \quad k\in\mathbb R$$

answered Oct 22 '19 at 19:32

Sebastiano

8,290

1

Please be more careful so not to reopen duplicates – Bill Dubuque Oct 31 '24 at 22:53
@BillDubuque For this question? I don't know of I have opened the duplicates. Excuse me. – Sebastiano Oct 31 '24 at 22:56
The question that I linked. If you don't have time to carefully assess a duplicate decision then please leave it to others - so we don't risk destroying site organization. – Bill Dubuque Oct 31 '24 at 22:59
@BillDubuque Oh, try to be more flexible and less rigid. I’m sure you never make mistakes. – Sebastiano Oct 31 '24 at 23:06

score 2 · Answer 3 · edited Oct 22 '19 at 19:46

Start by multiplying the top and bottom by $\sec(x)$, which will result in the following:

$$\int{\frac{\sec x}{\sec x-\tan x+1}}$$

Now we can perform a $u-$substitution for the following:

$$u=(\sec x-\tan x)\\du=(\sec x)(-(\sec x-\tan x))dx$$

The resulting integral is then:

$$\int{\frac{-1}{(u+1)(u)}}$$

We can now make use of partial fraction in the same manner that you did, giving us the following:

$$\int{\frac{1}{u+1}}+\int{\frac{-1}{u}}$$

Completing the remaining integrals and plugging back in for our substitution gives the following solution:

$$\ln\left|{\frac{1-\sin x+\cos x}{1-\sin x}}\right|+c$$

$\int \frac{dx}{1-\sin x+\cos x}$

3 Answers3