1. ## Integration problem

Hello everyone,
I've been preparing for my cal exam and I got stuck on this problem:
Evaluate the infinite integral
$\int \frac {\sec^4 x}{\sqrt(tan 2x)} dx$
I would really appreciate if you give me a clue on how I should do it because I've been trying different trigonometric identities and other methods for quite some time with no results.
Thank you

2. Originally Posted by karfag
Hello everyone,
I've been preparing for my cal exam and I got stuck on this problem:
Evaluate the infinite integral
$\int \frac {\sec^4 x}{\sqrt(tan 2x)} dx$
I would really appreciate if you give me a clue on how I should do it because I've been trying different trigonometric identities and other methods for quite some time with no results.
Thank you

3. I think you did not notice that it is a square root of tan(2x)... That is exactly what I was doing except that with square root it does not seem to lead anywhere.

4. Originally Posted by karfag
I think you did not notice that it is a square root of tan(2x)... That is exactly what I was doing except that with square root it does not seem to lead anywhere.
Edit- lag. You're right, give me a moment i'll try something

5. Hi everyone,

I don't think we can solve this integral using general methods. I mean it needs some advanced mathematics. Please refer, Wolfram Mathematica Online Integrator

6. Originally Posted by Sudharaka
Hi everyone,

I don't think we can solve this integral using general methods. I mean it needs some advanced mathematics. Please refer, Wolfram Mathematica Online Integrator

Holy bleeping ****, yeah i give up right now LOL

7. Eh, well I found it in previous final exam. First year college calculus.. I know the answer but I wanted to know how to get it. Well thanks for a try, if I'll figure it out, I'll post it lol

8. Originally Posted by karfag
Eh, well I found it in previous final exam. First year college calculus.. I know the answer but I wanted to know how to get it. Well thanks for a try, if I'll figure it out, I'll post it lol
Dear karfag,

(tan(2x))^1/2 + 1/5 * (tan(2x))^5/2 + C

Sorry for not writing it the correct way, I'm not very familiar with the math tags syntax

10. Originally Posted by karfag

(tan(2x))^1/2 + 1/5 * (tan(2x))^5/2 + C

Sorry for not writing it the correct way, I'm not very familiar with the math tags syntax
Dear karfag,

Differentiate $(tan(2x))^{1/2} + \frac{1}{5} (tan(2x))^{5/2}+C$ and you would obtain, $\frac {\sec^{4} 2x}{\sqrt{tan 2x}}$. Hence the integration should be $\int\frac {\sec^{4} 2x}{\sqrt{tan 2x}}$

11. If we let $u = \sqrt{\tan{2x}}$, then by the chain rule: $\dfrac{du}{dx} = \dfrac{\sec^2{2x}}{\sqrt{\tan{2x}}} \Rightarrow {dx} = \dfrac{\sqrt{\tan{2x}}}{\sec^2{2x}}\;{du}$. Thus $\int\dfrac{\sec^4{x}}{\sqrt{\tan{2x}}}\;{dx} = \int\left(\dfrac{\sec^4{x}}{\sqrt{\tan{2x}}}\right )\left(\dfrac{\sqrt{\tan{2x}}}{\sec^2{2x}}\right)\ ;{du} = \int\dfrac{\sec^4{x}}{\sec^2{2x}}\;{du}$. It gets bit complicated after that, though.

12. ... and since $\sec^4(x) = \left(1+\tan^2{x}\right)\left(1+\tan^2{x}\right)$ and $\sec^2{2x} = \left\{\dfrac{1+\tan^2{x}}{1-\tan^2{x}}\right\}^2,$ then $$\dfrac{\sec^4{x}}{\sec^2{2x}}$$ $= \dfrac{\left(1+\tan^2{x}\right)\left(1+\tan^2{x}\r ight)}{\left\{\dfrac{1+\tan^2{x}}{1-\tan^2{x}}\right\}^2}$ $= \dfrac{\left(1+\tan^2{x}\right)^2\left(1-\tan^2{x}\right)^2}{\left(1+\tan^2{x}\right)^2} = \left(1-\tan^2{x}\right)^2 = 1-2\tan^2{x}+\tan^4{x}$. Therefore $\int\dfrac{\sec^4{x}}{\sec^2{2x}}\;{du} = \sqrt{\tan{2x}}-2\int{\tan^2{x}}\;{du}+\int{\tan^4{x}}\;{du}
$
.

13. Originally Posted by TheCoffeeMachine
... and since $\sec^4(x) = \left(1+\tan^2{x}\right)\left(1+\tan^2{x}\right)$ and $\sec{2x} = \dfrac{1+\tan^2{x}}{1-\tan^2{x}},$ then $\color{red}{\dfrac{\sec^4{x}}{\sec{2x}} = \dfrac{\left(1+\tan{x}\right)\left(1+tan^2{x}\righ t)}{\left(\dfrac{1+\tan^{x}}{1-\tan^2{x}}\right)}}$ $= \dfrac{\left(1+\tan^2{x}\right)\left(1+tan^2{x}\ri ght)\left(1-\tan^2{x}\right)}{\left(1+\tan^2{x}\right)} = \left(1+\tan^2{x}\right) \left(1-\tan^2{x}\right) = 1-\tan^4{x}$. Therefore $\int\dfrac{\sec^4{x}}{\sec^2{2x}}\;{du} = \int{\left(1-\tan^4{x}\right)}\;{du} = u-\int{\tan^4{x}}\;{du} = \sqrt{\tan{2x}}-\int\tan^4{x}\;{du}.$
Dear TheCoffeeMachine,

There are some incorrect calculations which I have highlighted. Please recheck them. It should be,

$\frac{\sec^{4}x}{\sec^{2}2x} = \frac{\left(1+\tan^{2}x\right)\left(1+\tan^{2}x\ri ght)}{\left(\frac{1+\tan^{2}x}{1-\tan^{2}x}\right)^2}$

14. Originally Posted by karfag

(tan(2x))^1/2 + 1/5 * (tan(2x))^5/2 + C

Sorry for not writing it the correct way, I'm not very familiar with the math tags syntax

The integrand should be $\frac{ \sec^4(2x) }{\sqrt{ \tan(2x) }}$ .

If it is $\sec^4(x)$ , i believe that the integral cannot be expressed in terms of elementary function . I turn it into another one :

$\int \sqrt{ \frac{1}{x} - x }~dx$ but the definite integral

$\int_0^1 \sqrt{ \frac{1}{x} - x }~dx$ I think it should be in terms of $\Gamma(\frac{1}{4})$ .

15. Originally Posted by Sudharaka
There are some incorrect calculations which I have highlighted. Please recheck them.
Fixed. Thanks. It's become: $\sqrt{\tan{2x}}-2\int{\tan^2{x}}\;{du}+\int{\tan^4{x}}\;{du}$.