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Some Indefinite Integrals

jacks

Well-known member
Apr 5, 2012
226
[1] $\displaystyle \int\sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\cdot \frac{\sec x}{\sqrt{1+2\sec x}}dx$

[2] $\displaystyle \int \frac{3\cot 3x - \cot x}{\tan x-3 \tan 3x}dx$

Thanks pranav I have edited it.
 
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Pranav

Well-known member
Nov 4, 2013
428
[1] $\displaystyle \int\sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}\cdot \frac{\sec x}{\sqrt{1+2\sec x}}}dx$
I guess the forum requires to show some attempt on the problem.

For the first one, can you re-check with the source? I suspect that $\sec(x)/\sqrt{1+2\sec(x)}$ belongs outside the square root.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I guess the forum requires to show some attempt on the problem...
Yes, our goal is to help with problems, and when no work or thoughts are given, we can't really do that effectively. This does not mean however that you (or anyone) cannot give hints to get the OP started if you so desire. :D
 

Pranav

Well-known member
Nov 4, 2013
428
Yes, our goal is to help with problems, and when no work or thoughts are given, we can't really do that effectively. This does not mean however that you (or anyone) cannot give hints to get the OP started if you so desire. :D
Nice to know that we are allowed to give hints even when no work is shown. I will be posting some problems which I couldn't really attempt. I mean I have no idea where to begin with them so a push in the right direction would be greatly appreciated. :)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Nice to know that we are allowed to give hints even when no work is shown. I will be posting some problems which I couldn't really attempt. I mean I have no idea where to begin with them so a push in the right direction would be greatly appreciated. :)
In the event that you post a problem, and you are having trouble even beginning the problem, as long as you indicate this in your opening post, then our helpers know where to start with giving help. What we discourage is the posting of problems with no indication of where to begin giving help.
 

jacks

Well-known member
Apr 5, 2012
226
Sorry friends

Yes pranav it is outside the square root.

My approach for (I) one

$\displaystyle = \int\sqrt{\frac{1-\cos x}{1+\cos x}}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}dx$

$\displaystyle = \int\frac{1-\cos x}{\sin x}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}dx$

$\displaystyle = \int\frac{1}{1+\cos x}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}\cdot \sin xdx$

Now Let $\cos x = t^2$ Then $\sin xdx = -2tdt$

$\displaystyle = -\int\frac{1}{1+t^2}\cdot \frac{1}{\sqrt{t^2+2}}\cdot \frac{1}{t}\cdot 2tdt$

$\displaystyle = - 2\int\frac{1}{\left(1+t^2\right)\cdot \sqrt{t^2+2}}dt$

Now i have edited it.

Now Help me

For (II) one

$\displaystyle \int \frac{3\cot 3x - \cot x}{\tan x-3 \tan 3x}dx$

$\displaystyle \int\frac{3\tan 3x -\tan x}{\tan x- 3\tan 3x}\cdot \frac{1}{\tan x\cdot \tan 3x}dx$

$\displaystyle = -\int\cot x \cdot \cot 3xdx$

Now Help me

Thanks
 
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Pranav

Well-known member
Nov 4, 2013
428
Sorry friends

Yes pranav it is outside the square root.

My approach for (I) one

$\displaystyle = \int\sqrt{\frac{1-\cos x}{1+\cos x}}\cdot \frac{1}{\sqrt{2\cos x+1}\cdot \sqrt{\cos x}}dx$

$\displaystyle = \int\frac{1-\cos x}{\sin x}\cdot \frac{1}{\sqrt{2\cos x+1}\cdot \sqrt{\cos x}}dx$
Nope, this is incorrect. How do you get this?

Hint: Use $\cos(x)=2\cos^2(x/2)-1=1-2\sin^2(x/2)$.
For (II) one

$\displaystyle \int \frac{3\cot 3x - \cot x}{\tan x-3 \tan 3x}dx$

$\displaystyle \int\frac{3\tan 3x -\tan x}{\tan x- 3\tan 3x}\cdot \frac{1}{\tan x\cdot \tan 3x}dx$
Incorrect. Its $3\tan(x)-\tan(3x)$ in the numerator.

Next, can you express $\tan(3x)$ in terms of $\tan(x)$?
 

jacks

Well-known member
Apr 5, 2012
226
Thanks pranav.

My approach for (I) one

$\displaystyle = \int\sqrt{\frac{1-\cos x}{1+\cos x}}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}dx$

$\displaystyle = \int\frac{1-\cos x}{\sin x}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}dx$

$\displaystyle = \int\frac{1}{1+\cos x}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}\cdot \sin xdx$

Now Let $\cos x = t^2$ Then $\sin xdx = -2tdt$

$\displaystyle = -\int\frac{1}{1+t^2}\cdot \frac{1}{\sqrt{t^2+2}}\cdot \frac{1}{t}\cdot 2tdt$

$\displaystyle = - 2\int\frac{1}{\left(1+t^2\right)\cdot \sqrt{t^2+2}}dt$

Now Let $\displaystyle t = \frac{1}{u}$, Then $\displaystyle dt = -\frac{1}{u^2}du$

$\displaystyle = -\int\frac{u^3}{\left(1+u^2\right)\sqrt{1+2u^2}} \cdot\frac{1}{u^2}du = -\int\frac{u}{\left(1+u^2\right) \sqrt{2\left(u^2+1\right)-1}}du$

Now Let $u^2+1 = v^2$ , Then $2udu = 2vdv\Rightarrow udu = vdv$

Is it Right or not
 

jacks

Well-known member
Apr 5, 2012
226
Thanks pranav

$\displaystyle \int\frac{3\cot 3x -\cot x}{\tan x-3\tan 3x}dx = \int\frac{3\tan x-\tan 3x}{\tan x-3\tan 3x}\cdot \frac{1}{\tan x\cdot \tan 3x}dx$

Using $\displaystyle \tan 3x = \frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$

$\displaystyle = \int\frac{1-3\tan^2 x}{3-\tan^2 x}dx = \int\frac{3\left(3-\tan^2 x\right)-8}{3-\tan^2 x}dx$

$\displaystyle = 3\int 1dx -8\int\frac{1}{3-\tan^2x}dx$

Now Let $\tan x= t$ and $\displaystyle \sec^2 xdx = dt\Rightarrow dx = \frac{1}{1+\tan^2 x}dt = \frac{1}{1+t^2}dt$

$\displaystyle = 3x-8\int\frac{1}{\left(1+t^2\right)\cdot \left(3-t^2\right)}dt = 3x+8\int\frac{1}{\left(1+t^2\right)\cdot \left(t^2-3\right)}dt$

$\displaystyle = 3x+8\cdot \frac{1}{4}\int\frac{1}{t^2-3}dt-\frac{8}{4}\int\frac{1}{1+t^2}dt$

$\displaystyle = 3x+2\cdot \ln \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|-2\cdot \tan^{-1}\left(t\right)+\mathbb{C}$

$\displaystyle = 3x+2\cdot \ln \left|\frac{\tan x-\sqrt{3}}{\tan x+\sqrt{3}}\right|-2\cdot \tan^{-1}\left(\tan x\right)+\mathbb{C}$
 

Pranav

Well-known member
Nov 4, 2013
428
Thanks pranav.

My approach for (I) one

$\displaystyle = \int\sqrt{\frac{1-\cos x}{1+\cos x}}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}dx$

$\displaystyle = \int\frac{1-\cos x}{\sin x}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}dx$
Not correct still.

$$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}} \neq \frac{1-\cos(x)}{\sin(x)}$$

I don't see how you get this.

Thanks pranav

$\displaystyle \int\frac{3\cot 3x -\cot x}{\tan x-3\tan 3x}dx = \int\frac{3\tan x-\tan 3x}{\tan x-3\tan 3x}\cdot \frac{1}{\tan x\cdot \tan 3x}dx$

Using $\displaystyle \tan 3x = \frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$
Good! ;)

$\displaystyle = \int\frac{1-3\tan^2 x}{3-\tan^2 x}dx = \int\frac{3\left(3-\tan^2 x\right)-8}{3-\tan^2 x}dx$

$\displaystyle = 3\int 1dx -8\int\frac{1}{3-\tan^2x}dx$

Now Let $\tan x= t$ and $\displaystyle \sec^2 xdx = dt\Rightarrow dx = \frac{1}{1+\tan^2 x}dt = \frac{1}{1+t^2}dt$

$\displaystyle = 3x-8\int\frac{1}{\left(1+t^2\right)\cdot \left(3-t^2\right)}dt = 3x+8\int\frac{1}{\left(1+t^2\right)\cdot \left(t^2-3\right)}dt$

$\displaystyle = 3x+8\cdot \frac{1}{4}\int\frac{1}{t^2-3}dt-\frac{8}{4}\int\frac{1}{1+t^2}dt$

$\displaystyle = 3x+2\cdot \ln \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|-2\cdot \tan^{-1}\left(t\right)+\mathbb{C}$

$\displaystyle = 3x+2\cdot \ln \left|\frac{\tan x-\sqrt{3}}{\tan x+\sqrt{3}}\right|-2\cdot \tan^{-1}\left(\tan x\right)+\mathbb{C}$
Looks very good to me. Great job! :)
 

jacks

Well-known member
Apr 5, 2012
226
To Pranav

$\displaystyle \sqrt{\frac{1-\cos x}{1+\cos x}} = \sqrt{\frac{1-\cos x}{1+\cos x}\times \frac{1-\cos x}{1-\cos x}} = \sqrt{\frac{\left(1-\cos x\right)^2}{\sin^2 x}} = \left|\frac{1-\cos x}{\sin x}\right| = \frac{1-\cos x}{\sin x}$ for $x\in \left(0,\pi\right)$
 

Pranav

Well-known member
Nov 4, 2013
428
To Pranav

$\displaystyle \sqrt{\frac{1-\cos x}{1+\cos x}} = \sqrt{\frac{1-\cos x}{1+\cos x}\times \frac{1-\cos x}{1-\cos x}} = \sqrt{\frac{\left(1-\cos x\right)^2}{\sin^2 x}} = \left|\frac{1-\cos x}{\sin x}\right| = \frac{1-\cos x}{\sin x}$ for $x\in \left(0,\pi\right)$
I am very sorry to mislead you. I posted my reply in the morning when I was in hurry to leave for the tuition. I realised it on my way there. I am going to check your work again.

Very sorry again.
 

Pranav

Well-known member
Nov 4, 2013
428
Okay, your approach for 1 in #8 looks good so far but I suspect you will have to use one more substitution after the last substitution you have presented. What I had in my mind was the following:

Rewrite:
$$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\tan(x/2)$$
$$\sqrt{\cos(x)}=\sqrt{2\cos^2(x/2)-1}$$
$$\sqrt{\cos(x)+2}=\sqrt{2\cos^2(x/2)+1}$$

So we have the following integral:
$$\int \frac{\tan(x/2)}{\sqrt{2\cos^2(x/2)-1}\sqrt{2\cos^2(x/2)+1}}dx=\int \frac{\tan(x/2)}{\sqrt{4\cos^4(x/2)-1}}dx$$
$$=\int \frac{\tan(x/2)\sec^2(x/2)}{\sqrt{4-\sec^4(x/2)}}dx$$
Now use the substitution, $\sec^2(x/2)=t \Rightarrow \sec^2(x/2)\tan(x/2)dx=dt$.
Hence, our integral transforms to:
$$\int \frac{dt}{\sqrt{4-t^2}}$$

The above integral is quite straightforward.

I hope that helped. :)