Integrate $\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$

Question

Integrate: $$\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$$

My attempt :

I substituted $\sec{x} + \tan{x} $ as $t^2$

Then, $$ (\sec{x} \cdot \tan{x} + \sec^2x) dx =2tdt$$

$$\sec{x}( \tan{x} + \sec{x}) dx =2tdt$$

$$\sec{x}\cdot t^2 dx =2tdt$$

But there is $\sec{x}$ left outside.

I thought about this possibility because of your problem with $dx$. I shall look at it now. — Claude Leibovici, Aug 19 '14 at 03:28

score 5 · Accepted Answer · answered Aug 19 '14 at 03:32

$\bf{My\; Solution::}$ Let $\displaystyle I = \int \sqrt{\sec x+\tan x}\cdot \sec^2 xdx = \int\frac{\sqrt{\sec^2 x-\tan^2 x}}{\sqrt{\sec x-\tan x}}\cdot \sec^2 xdx$

Now Integral $\displaystyle I = \int\frac{\sec^2 x}{\left(\sec x-\tan x \right)^{\frac{1}{2}}}dx$

Now Let $\left(\sec x-\tan x \right) = t^2\;,$ Then $\sec x\cdot \left(\tan x-\sec x\right)dx = 2tdt\Rightarrow \sec x\cdot t dx =-2tdt$

So $\sec xdx = -2dt$ So Integral $\displaystyle I = -2\int \frac{\sec x }{t}dt$

Now Using $\displaystyle \left(\sec x- \tan x\right) = t$ and $\displaystyle \left(\sec x+\tan x \right) = \frac{1}{t}$

Using $\left(\sec^2 x-\tan^2 x\right) = 1$

So $\displaystyle 2\sec x = t+\frac{1}{t} = \frac{t^2+1}{t}$

So Integral $\displaystyle I = -\int\frac{t^2+1}{t^2}dt = -t+\frac{1}{t}+\mathbb{C}$

So $\displaystyle I = \int \sqrt{\sec x+\tan x}\cdot \sec^2 xdx = -\left(\sec x-\tan x\right)+\frac{1}{\left(\sec x -\tan x\right)}+\mathbb{C}$

$\displaystyle = -\left(\sec x-\tan x\right)+\left(\sec x+\tan x\right)+\mathbb{C} = 2\tan x+\mathbb{C}$

@mathslover. I think I agree with you but there is an idea to explore. — Claude Leibovici, Aug 19 '14 at 03:37
Method is correct...! the correction is $\displaystyle \left(\sec x- \tan x\right) = t^2$ — maths lover, Aug 19 '14 at 03:38
I have a problem with your answer (I am unable to provide one !). But $$\int_0^{\pi/4} \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x,dx=\frac{2}{3} \left(1+\sqrt{2 \left(\sqrt{2}-1\right)}\right)$$ — Claude Leibovici, Aug 19 '14 at 04:27

score 1 · Answer 2 · answered Aug 19 '14 at 03:50

Hmmm... It's possible double angle identities will work here.

$$\int\sqrt{\sec x+\tan x}\sec^2xdx=\int\sec^{5/2}x\sqrt{1+\sin x}dx=$$ $$\int\sec^{5/2}x\sqrt{\sin^2\frac x2+\cos^2\frac x2+2\sin\frac x2\cos\frac x2}dx=$$ $$\int(\cos^2\frac x2-\sin^2 \frac x2)^{5/2}(\sin\frac x2+\cos\frac x2)dx=$$ $$\int(\cos\frac x2-\sin\frac x2)^{5/2}(\cos\frac x2+\sin\frac x2)^{7/2}dx$$

Not sure where to go from here. Maybe someone else can find a way to make this work.

Claude Leibovici · Answer 3 · 2014-08-19T06:17:13.840

If we use Weierstrass substitution, the problem becomes$$I=\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx=2 \int \frac{ \sqrt{\frac{1+t}{1-t}} \left(1+t^2\right)}{\left(1-t^2\right)^2}\,dt$$ Now, setting $$\sqrt{\frac{1+t}{1-t}}=z$$ the integral becomes $$I=\int (z^2+\frac{1}{z^2})\,dz=\frac{z^3}{3}-\frac{1}{z}$$ Now, back to $x$, we get $$I=\frac{2}{3} \Big(2 \sin (x)-1\Big) \sec (x) \sqrt{\frac{\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}}$$

Integrate $\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$

3 Answers3