How do you integrate \(\int \sec^{2n+1} x \, dx\)?

[nhrock3]

\intop_{-\pi/2}^{+\pi/2}\sqrt{(-6sin2t)^2+\sqrt{6}(cost)^2}dt

how to solve it?
i tried
by parts and it looks very bad and complicated

edit
sorry i copied the wrond integral
now it fine

 

[mateomy]

Did you try a U substitution first?

 

[nhrock3]

what substitution?

 

[mateomy]

A "u" substitution. Where you pick a section of your integral and substitute it with "u" and replace your limits of integration. Its not in your book?

 

[nhrock3]

ok ill take u=sint
so du=costdt
but i can't have dt inside the root
and i can't take it aside apart

 

[mateomy]

You're going to need to use a trig identity too. Sorry, I am doing this in my head.

 

[nhrock3]

what identity?

 

[mateomy]

Look at your first term -12sintcost it should be pretty obvious after that.

 

[nhrock3]

yes i know it wwas sin2t before
and i brke it into 2sintcost
what to do?

 

[nhrock3]

i did z=sint
and simplified it to
\intop_{-\pi/2}^{+\pi/2}\sqrt{36z^2+6}dz

what now?

 

[hunt_mat]

1) You also have to also change the limits of integration, I think they are -1 and 1 now.

2) Take out a factor of 36 from the integral

3) This should now be a standard integral. If you need more assistance, I would think of letting z=\sinh w/\sqrt{6} and working from there.

 

[nhrock3]

i haven't studied hiporbolic functions
could you write it into normal substitution
?

 

[hunt_mat]

Okay, try z=\tan w/\sqrt{6}

 

[Mark44]

Look at your first term -12sintcost it should be pretty obvious after that.

Maybe I'm forgetting something, but it's not obvious to me what to do after that.

i did z=sint
and simplified it to
\intop_{-\pi/2}^{+\pi/2}\sqrt{36z^2+6}dz

what now?

I don't see how you got this from your substitution of z = sin(t). dz = cos(t)dt, and as you pointed out already, you don't have a factor of cos(t) outside the integral.

 

[nhrock3]

Okay, try z=\tan w/\sqrt{6}

i did it
i got dz/(cos w) type integral
but i can't substitute dz with a w variable
?

 

[Mark44]

i did z=sint
and simplified it to
\intop_{-\pi/2}^{+\pi/2}\sqrt{36z^2+6}dz

what now?

1) You also have to also change the limits of integration, I think they are -1 and 1 now.

2) Take out a factor of 36 from the integral

3) This should now be a standard integral. If you need more assistance, I would think of letting z=\sinh w/\sqrt{6} and working from there.

I could be wrong, but in step 3, hunt_mat is taking the integral at the top at face value. I don't believe that this integral follows from the original integral of this thread.

 

[hunt_mat]

I could be wrong, but in step 3, hunt_mat is taking the integral at the top at face value. I don't believe that this integral follows from the original integral of this thread.

Correct I was.

 

[nhrock3]

he is following the right integral
it developed into this new one
https://www.physicsforums.com/showpost.php?p=3247073&postcount=10
and here i tried to solve it
https://www.physicsforums.com/showpost.php?p=3247106&postcount=15

 

[hunt_mat]

The integral boils down to integrating

<br /> \int\sqrt{1+x^{2}}dx<br />

though.

 

[nhrock3]

ok i will try this thing tommorow
good night :)

 

[Mark44]

\intop_{-\pi/2}^{+\pi/2}\sqrt{(-6sin2t)^2+\sqrt{6}(cost)^2}dt

how to solve it?
i tried
by parts and it looks very bad and complicated

edit
sorry i copied the wrond integral
now it fine

Well, that makes a big difference. As I recall, you originally had this integral.

\int_{-\pi/2}^{+\pi/2}\sqrt{-6sin2t+\sqrt{6}cost}dt

 

[SammyS]

Yes Mark. That editing feature can jump up and bite you.

 

[nhrock3]

i got now an integral of dt/(cost)^3

what to do?

 

[hunt_mat]

I am unsure how to do this integral without using hyperbolic function, I used the following functions:

<br /> \begin{array}{rcl}<br /> \sinh x &amp; = &amp; \frac{e^{x}-e^{-x}}{2} \\<br /> \cosh x &amp; = &amp; \frac{e^{x}+e^{-x}}{2} \\<br /> \cosh^{2}x-\sinh^{2}x &amp; \equiv &amp; 1 \\<br /> \cosh 2x &amp; = &amp; \cosh^{2}x+\sinh^{2}x \\<br /> \sinh 2x &amp; = &amp; 2\sinh x\cosh x<br /> \end{array}<br />

With the above the integral becomes simple. The first two are definitions of cosh and sinh, the rest may be derived from those definitions.

 

[nhrock3]

how it become simple?

 

[hunt_mat]

For the integral I posted:

<br /> \int\sqrt{1+x^{2}}dx<br />

Use the substitution x=\sinh u and what do you get?

 

[nhrock3]

i used x=tan t substitution
and i got a different integral
dt/(cos t)^3
how to solve it

 

[hunt_mat]

I am not sure you can do it that way anymore, as I said, I have been thinking and the only straightforward way of doing it is hyperbolic functions, I put everything that you need in a previous post.

 

[nhrock3]

i want to solve it my way
i got now an integral of dt/(cost)^3

what to do?

 

[hunt_mat]

The only thing that I can think of is use:

<br /> t=\tan\left(\frac{x}{2}\right)<br />

I don't think that it will get you anywhere with this method, good luck though.

 

[nhrock3]

ok but i get the same integral s before

 

[hunt_mat]

Like I said, I can't think of any other way of doing it other than the way I have mentioned. Perhaps your lecturer can.

 

[Gib Z]

Integrating I_n = \int \sec^{2n+1} x dx:

1. Integrate by parts to get a recurrence relation.
With u=\sec^{2n-1} x , du = (2n-1) \sec^{2n-2} x \cdot \sec x \tan x dx and dv = \sec^2 x dx, v= \tan x we have
I_n = \int \sec^{2n+1} x dx = \sec^{2n-1} x \tan x - \int (2n-1) \sec^{2n-1}\cdot \tan^2 dx
= \sec^{2n-1} x \tan x - (2n-1) \biggl( \int sec^{2n+1}x dx- \int \sec^{2n-1}x dx \biggr)
= \sec^{2n-1} x \tan x -(2n-1)\left( I_n - I_{n-1} \right)
I_n = \frac{\sec^{2n-1} x \tan x +(2n-1)I_{n-1}}{2n}

This eventually reduces the problem down into I_1 = \int \sec x dx = \int \frac{\sec x (\sec x + \tan x)}{\sec x + \tan x} dx = \log |\sec x + \tan x| + C

2. Make a substitution, then integrate a rational function.
I_n = \int \frac{\cos x}{\cos^{2n+2} x}{dx} = \int \frac{ d(\sin x) }{(1-\sin^2 x)^{n+1} } = \int \frac{1}{(1-u^2)^{n+1}} du
Now it is a simple but tedious exercise in partial fractions.

3. Let x= \sin^{-1}\left(\tanh t\right), then
I_n = \int \cosh^{2n} t dt

\cosh^{2n} x= \left( \frac{e^x + e^{-x} }{2} \right)^{2n}
= \frac{1}{2^{2n}}\left( \binom{n}{0} e^{2nx} + \binom{n}{1} e^{2(n-1)x} + \binom{n}{2} e^{2(n-2)x} + \cdots + \binom{n}{n-2} e^{-2(n-2)x} + \binom{n}{n-1} e^{-2(n-1)x} + \binom{n}{n} e^{-2nx} \right)
= \frac{1}{2^{2n-1}} \left( \binom{n}{0} \cosh (2nx) + \binom{n}{1} \cosh (2(n-1)x) + \binom{n}{2} \cosh ( 2(n-2)x ) + \cdots

So integrating term by term:
I_n = \frac{1}{2^{2n}} \left( \binom{n}{0} \frac{\sinh (2nt)}{n} + \binom{n}{1} \frac{\sinh (2(n-1)t) }{n-1} + \binom{n}{2} \frac{ \sinh (2(n-2)t)}{n-2} + \cdots + C
where t= \tanh^{-1} \left(\sin x\right).