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I guess the forum requires to show some attempt on the problem.[1] $\displaystyle \int\sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}\cdot \frac{\sec x}{\sqrt{1+2\sec x}}}dx$
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.I guess the forum requires to show some attempt on the problem...
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.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.![]()
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.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.![]()
Nope, this is incorrect. How do you get this?Sorry friends
Yes pranav it is outside the square root.
My approach for (I) one
$\displaystyle = \int\sqrt{\frac{1-\cos x}{1+\cos x}}\cdot \frac{1}{\sqrt{2\cos x+1}\cdot \sqrt{\cos x}}dx$
$\displaystyle = \int\frac{1-\cos x}{\sin x}\cdot \frac{1}{\sqrt{2\cos x+1}\cdot \sqrt{\cos x}}dx$
Incorrect. Its $3\tan(x)-\tan(3x)$ in the numerator.For (II) one
$\displaystyle \int \frac{3\cot 3x - \cot x}{\tan x-3 \tan 3x}dx$
$\displaystyle \int\frac{3\tan 3x -\tan x}{\tan x- 3\tan 3x}\cdot \frac{1}{\tan x\cdot \tan 3x}dx$
Not correct still.Thanks pranav.
My approach for (I) one
$\displaystyle = \int\sqrt{\frac{1-\cos x}{1+\cos x}}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}dx$
$\displaystyle = \int\frac{1-\cos x}{\sin x}\cdot \frac{1}{\sqrt{\cos x+2}\cdot \sqrt{\cos x}}dx$
Good!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 $\displaystyle \tan 3x = \frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$
Looks very good to me. Great job!$\displaystyle = \int\frac{1-3\tan^2 x}{3-\tan^2 x}dx = \int\frac{3\left(3-\tan^2 x\right)-8}{3-\tan^2 x}dx$
$\displaystyle = 3\int 1dx -8\int\frac{1}{3-\tan^2x}dx$
Now Let $\tan x= t$ and $\displaystyle \sec^2 xdx = dt\Rightarrow dx = \frac{1}{1+\tan^2 x}dt = \frac{1}{1+t^2}dt$
$\displaystyle = 3x-8\int\frac{1}{\left(1+t^2\right)\cdot \left(3-t^2\right)}dt = 3x+8\int\frac{1}{\left(1+t^2\right)\cdot \left(t^2-3\right)}dt$
$\displaystyle = 3x+8\cdot \frac{1}{4}\int\frac{1}{t^2-3}dt-\frac{8}{4}\int\frac{1}{1+t^2}dt$
$\displaystyle = 3x+2\cdot \ln \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|-2\cdot \tan^{-1}\left(t\right)+\mathbb{C}$
$\displaystyle = 3x+2\cdot \ln \left|\frac{\tan x-\sqrt{3}}{\tan x+\sqrt{3}}\right|-2\cdot \tan^{-1}\left(\tan x\right)+\mathbb{C}$
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.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 $x\in \left(0,\pi\right)$