Some Indefinite Integrals

[jacks]

[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.

 

[Pranav]

[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]

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.

 

[Pranav]

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.

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]

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]

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

 

[Pranav]

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]

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]

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]

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]

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]

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]

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.