# Thread: Evaluate the integral : secx/sqrt(sin2x)

1. ## Evaluate the integral : secx/sqrt(sin2x)

Hello
again i got i headache ;p

Evaluate the integral:
secx/sqrt(sin2x)

sin(2x) not sin^2(x)

2. Originally Posted by TWiX
Hello
again i got i headache ;p

Evaluate the integral:
secx/sqrt(sin2x)

sin(2x) not sin^2(x)
Mathematica gives the answer as $\sec(x) \sqrt{\sin(2x)}$

Since the derivative of this would involve more than one term, I'm thinking integration by parts a logical choice but let me think some more about it. What have you tried?

3. $\frac{1}{\cos(x) \sqrt{\sin(2x)}}$

$= \frac{1 + \cos(2x) - \cos(2x) }{\cos(x) \sqrt{\sin(2x)}}$

$= \frac{2\cos(x)}{\sqrt{\sin(2x)}} - \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}}$

its integral $=$

$\int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \int \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}} ~dx$

$= \int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \sqrt{\sin(2x)}\sec(x) + \int \sqrt{\sin(2x)} \sec(x) \tan(x) ~dx$

$= \int \frac{ 2\cos^2(x) + 2\sin^2(x)}{ \cos(x) \sqrt{\sin(2x)} } ~dx - \sqrt{\sin(2x)}\sec(x) +C$

therefore

$I = 2I - \sqrt{\sin(2x)}\sec(x) + C ~ , ~ I = \sqrt{\sin(2x)}\sec(x) + C'$

It is just a trial ... but luckily it is a successful trial

4. Originally Posted by Jameson
Mathematica gives the answer as $\sec(x) \sqrt{\sin(2x)}$

Since the derivative of this would involve more than one term, I'm thinking integration by parts a logical choice but let me think some more about it. What have you tried?
i tried by parts with u=secx
but it was hard
i tried to simplify it using trigonometric identities but it does not work

5. Originally Posted by simplependulum
$\frac{1}{\cos(x) \sqrt{\sin(2x)}}$

$= \frac{1 + \cos(2x) - \cos(2x) }{\cos(x) \sqrt{\sin(2x)}}$

$= \frac{2\cos(x)}{\sqrt{\sin(2x)}} - \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}}$

its integral $=$

$\int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \int \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}} ~dx$

$= \int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \sqrt{\sin(2x)}\sec(x) + \int \sqrt{\sin(2x)} \sec(x) \tan(x) ~dx$

$= \int \frac{ 2\cos^2(x) + 2\sin^2(x)}{ \cos(x) \sqrt{\sin(2x)} } ~dx - \sqrt{\sin(2x)}\sec(x) +C$

therefore

$I = 2I - \sqrt{\sin(2x)}\sec(x) + C ~ , ~ I = \sqrt{\sin(2x)}\sec(x) + C'$

It is just a trial ... but luckily it is a successful trial
(1+cos(2x))/cosx = cosx??

6. Very nice job, simplependulum!

Originally Posted by TWiX
(1+cos(2x))/cosx = cosx??
I don't quite follow where you are point to. Do you mean the line (1+cos(2x)-cos(2x)), which is the second line in simplependulum's post?

7. Originally Posted by Jameson
Very nice job, simplependulum!

I am just wondering if there is someone who can think of

$1 = \tan(x) \sin(2x) + \cos(2x)$ ( the numberator obtained when we differentiate $\sec(x) \sqrt{\sin(2x)}$)

... i am sure that this guy is a genius !

8. Originally Posted by Jameson
Very nice job, simplependulum!

I don't quite follow where you are point to. Do you mean the line (1+cos(2x)-cos(2x)), which is the second line in simplependulum's post?
no
i know its = adding zero

am talking about the first fraction in the third line

9. Originally Posted by TWiX
no
i know its = adding zero

am talking about the first fraction in the third line
$\frac{1+cos(2x)}{cos(x)}$

$=\frac{1+2cos ^2(x)-1}{cos(x)}$

$=2 \cos(x)$

Originally Posted by simplependulum
I am just wondering if there is someone who can think of

$1 = \tan(x) \sin(2x) + \cos(2x)$ ( the numberator obtained when we differentiate $\sec(x) \sqrt{\sin(2x)}$)

... i am sure that this guy is a genius !
I'm sure you are genius !!

10. Here is another method :

$\int \frac{\sec(x)}{\sqrt{\sin(2x)}}~dx$

$= \int \frac{\sec^2(x) \cos(x) }{\sqrt{\sin(2x)}}~dx$

$= \int \sec^2(x) \frac{ \cos(x) }{ \sqrt{2} \sqrt{\sin(x)} \sqrt{\cos(x)}}~dx$

$= \frac{1}{\sqrt{2}} \int \sec^2(x) \frac{ 1}{\sqrt{\tan(x)}}~dx$

by substituting $t = \tan(x)$

i obtain

$I = \frac{1}{\sqrt{2}}2 \sqrt{\tan(x)} + C$

$I = \sqrt{2 \sin(x) \cos(x)} \sec(x) + C$