Integration problem

**karfag** · May 23rd 2010, 03:21 PM

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

**AllanCuz** · May 23rd 2010, 04:17 PM

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

Midread the integrand...

**karfag** · May 23rd 2010, 04:23 PM

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.

**AllanCuz** · May 23rd 2010, 04:49 PM

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

**Sudharaka** · May 23rd 2010, 04:58 PM

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

Hope this will help you.

**AllanCuz** · May 23rd 2010, 04:59 PM

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

Hope this will help you.

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

**karfag** · May 23rd 2010, 05:06 PM

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

**Sudharaka** · May 23rd 2010, 05:11 PM

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,

Did your answer tally with the answer "Wolfram" suggested??

**karfag** · May 23rd 2010, 05:16 PM

The answer listed is

(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

**Sudharaka** · May 23rd 2010, 07:07 PM

Originally Posted by karfag

The answer listed is

(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 $\int\frac {\sec^{4} 2x}{\sqrt{tan 2x}}$

**TheCoffeeMachine** · May 23rd 2010, 09:56 PM

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.

**TheCoffeeMachine** · May 24th 2010, 02:45 AM

... 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 [tex]\dfrac{\sec^4{x}}{\sec^2{2x}}[/Math] $= \dfrac{\left(1+\tan^2{x}\right)\left(1+\tan^2{x}\r ight)}{\left\{\dfrac{1+\tan^2{x}}{1-\tan^2{x}}\right\}^2}$ $= \dfrac{\left(1+\tan^2{x}\right)^2\left(1-\tan^2{x}\right)^2}{\left(1+\tan^2{x}\right)^2} = \left(1-\tan^2{x}\right)^2 = 1-2\tan^2{x}+\tan^4{x}$ . Therefore $\int\dfrac{\sec^4{x}}{\sec^2{2x}}\;{du} = \sqrt{\tan{2x}}-2\int{\tan^2{x}}\;{du}+\int{\tan^4{x}}\;{du}<br />$ .

**Sudharaka** · May 24th 2010, 03:52 AM

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,

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

Hope this will help you.

**simplependulum** · May 24th 2010, 06:41 AM

Originally Posted by karfag

The answer listed is

(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 $\Gamma(\frac{1}{4})$ .

**TheCoffeeMachine** · May 24th 2010, 08:44 AM

Originally Posted by Sudharaka

There are some incorrect calculations which I have highlighted. Please recheck them.

Fixed. Thanks. It's become: $\sqrt{\tan{2x}}-2\int{\tan^2{x}}\;{du}+\int{\tan^4{x}}\;{du}$ .

Thread: Integration problem

LinkBack

Thread Tools

Display

Integration problem

Similar Math Help Forum Discussions

integration problem

Simple Integration problem: Trigonometric Integration

Integration Problem

Integration problem !!

Integration problem

Search Tags