Evaluation of Indefinite Integral

[juantheron]

Calculation of \(\displaystyle \int\frac{\sqrt{\cos 2x}}{\sin x}dx\)

**My Try \(\displaystyle :: I = \int\frac{\sqrt{\cos 2x}}{\sin x}dx = \int\frac{\cos 2x}{\sin x\cdot \sqrt{\cos 2x}}\cdot \frac{\sin x}{\sin x}dx = \int\frac{\sqrt{2\cos^2 x-1}}{\left(1-\cos^2 x\right)\cdot \sqrt{2\cos^2 x-1}}\cdot \sin x dx\)Now Let \(\displaystyle \cos x = t\) and \(\displaystyle \sin xdx = -dt\)So Integral \(\displaystyle I = \int\frac{\left(2t^2-1\right)}{(t^2-1)\cdot \sqrt{2t^2-1}}dt\)

Now I did not Understand How can i proceed further.

Help me for solving above Question.

Thanks

 

[Euge]

Calculation of \(\displaystyle \int\frac{\sqrt{\cos 2x}}{\sin x}dx\)

**My Try \(\displaystyle :: I = \int\frac{\sqrt{\cos 2x}}{\sin x}dx = \int\frac{\cos 2x}{\sin x\cdot \sqrt{\cos 2x}}\cdot \frac{\sin x}{\sin x}dx = \int\frac{\sqrt{2\cos^2 x-1}}{\left(1-\cos^2 x\right)\cdot \sqrt{2\cos^2 x-1}}\cdot \sin x dx\)Now Let \(\displaystyle \cos x = t\) and \(\displaystyle \sin xdx = -dt\)So Integral \(\displaystyle I = \int\frac{\left(2t^2-1\right)}{(t^2-1)\cdot \sqrt{2t^2-1}}dt\)

Now I did not Understand How can i proceed further.

Help me for solving above Question.

Thanks

You're on the right track. Let's continue from where you left off. Break up the integral as

$\displaystyle 2\int \frac{dt}{\sqrt{2t^2-1}} + \int \frac{dt}{(t^2-1)\sqrt{2t^2-1}}$.

The first integral is

$\displaystyle \sqrt{2}\log|\sqrt{2}t + \sqrt{2t^2-1}| + C$,

which can be obtained by using the $u$-substitution $u = \cosh^{-1}(\sqrt{2} t)$.

The second integral takes more work. Using the partial fraction decomposition

$\displaystyle \frac{2}{t^2-1} = \frac{1}{t-1} - \frac{1}{t+1}$

we write the second integral as

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

To evalute these integrals, we'll consider a general quadratic $R(u) = R_{a, b, c}(u) := au^2 + bu + c$ and use the formula

$\displaystyle (*) \int \frac{du}{u\sqrt{R}} = -\frac{1}{\sqrt{c}}\log\left|\frac{\sqrt{R} + \sqrt{c}}{u} + \frac{b}{2\sqrt{c}}\right| + C, \quad \text{$c > 0$} $.

This formula may be derived by using the substitution $v = \frac{1}{u}$ and the formula

$\displaystyle \int \frac{dv}{\sqrt{R_{c, b, a}}} = \frac{1}{\sqrt{c}}\log|2cv + b + 2\sqrt{c}\sqrt{R_{c, b, a}}| + C$.

With formula (*) at hand, we evaluate

$\displaystyle \int \frac{dt}{(t-1)\sqrt{2t^2-1}}$

$\displaystyle = -\log\left|\frac{\sqrt{R_{1,2,1/2}(t-1)} + \sqrt{\frac{1}{2}}}{t-1} + \sqrt{2}\right| + C$

$\displaystyle = -\log\left|\frac{\sqrt{t^2-\frac{1}{2}} + \sqrt{\frac{1}{2}}}{t-1} + \sqrt{2}\right| + C$

$\displaystyle = -\log\left|\frac{\sqrt{2t^2-1}+2t-1}{t-1}\right| + C$.

By a similar analysis,

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

Thus

$\displaystyle \int \frac{dt}{(t^2-1)\sqrt{2t^2-1}}$

$\displaystyle = \frac{1}{2}\log\left|\frac{\sqrt{2t^2-1}-2t-1}{\sqrt{2t^2-1}-2t+1}\right| - \frac{1}{2}\log\left|\frac{t+1}{t-1}\right| + C$

$\displaystyle =\frac{1}{2}\log\left|\frac{\sqrt{\cos(2x)} - 2\cos(x) - 1}{\sqrt{\cos(2x)} - 2\cos(x) + 1}\right| + \frac{1}{2}\log\left|\frac{\cos(x) + 1}{\cos(x) - 1}\right|+C$.

At last, we obtain

$\displaystyle \int \frac{\sqrt{\cos(2x)}}{\sin(x)}\, dx$

$\displaystyle = \sqrt{2}\log|\sqrt{2}\cos(x) + \cos(2x)| + \frac{1}{2}\log\left|\frac{\sqrt{\cos(2x)} - 2\cos(x) - 1}{\sqrt{\cos(2x)} - 2\cos(x) + 1}\right| - \frac{1}{2}\log\left|\frac{\cos(x) + 1}{\cos(x) - 1}\right| + C$.

 

[Euge]

I just wanted to remark that in the expression just above the phrase "At last, we obtain," there should be an $=$ sign to the left and a $+C$ to the right.

 

[MarkFL]

I just wanted to remark that in the expression just above the phrase "At last, we obtain," there should be an $=$ sign to the left and a $+C$ to the right.

I just had to fix such a beautiful post. :D