Integration help...

[hotcommodity]

So, my math professor assigned homework over the break on material he didn't cover in class and that is not in our text book. Any help would be appreciated...

 

Ok, we're supposed to use basic trig identities to get the integral down to something manageable where we can use u-substitution and/or integration by parts. I've worked out one but I'm having trouble on the other two...

 

1. the integral of (cos(x))^4 dx. [Hint: also (cos(x)^2)^2]

 

2. the integral of sec(v) dv. [Hint: multiply sec(v) by (sec(v)+tan(v))/(sec(v)+tan(v)).]

 

Some identities to use:

 

sin(2x)=2sin(x)cos(x)

 

cos(2x)=(cos(x))^2-(sin(x))^2

 

cos(2x)=2(cos(x))^2-1

 

cos(2x)=1-2(sin(x))^2

 

(sin(x))^2=(1/2)[1-COS(2X)]

 

I'm not looking for the answer, just a push in the right direction, thanks. (Sorry about my lack of LaTex skills =/. )

[CPL.Luke]

well for the first one look at what identities include cosx ^2

[Klaynos]

For the first one, considering the identify:

 

cos(2x)=2(cos(x))^2-1

 

might prove advantageous...

[hotcommodity]

Ok, I tried using the identity cos(x) squared is equal to (1/2)*(1+cos(2x)). So then I have one half times the integral of (1+cos(2x))*(cos(x))^2. Then I tried plugging in 1-2(sin(x))^2 for cos(2x), but that brought me back to square one. Then I tried plugging in (cos(x))^2-(sin(x))^2 for cos(2x), but I'm not sure I can do anything with that. Am I headed down the right path?...

[EvoN1020v]

[math]cos^2(x)=1-sin^2(x)[/math]

 

You should know that [math]\int cosx = sinx + C[/math]

 

You need to know the u-substitution also.

[Klaynos]

Ok, I tried using the identity cos(x) squared is equal to (1/2)*(1+cos(2x)). So then I have one half times the integral of (1+cos(2x))*(cos(x))^2. Then I tried plugging in 1-2(sin(x))^2 for cos(2x), but that brought me back to square one. Then I tried plugging in (cos(x))^2-(sin(x))^2 for cos(2x), but I'm not sure I can do anything with that. Am I headed down the right path?...

 

(1+cos(2x))*(cos(x))^2.

 

From here I'd substitue the indentity for cosx^2 again, then probably multiply out the terms and do it again I think from that point it's alot easier. Also I think there may be a problem in your substitution, a /2 missing possibly.

[EvoN1020v]

I'm not sure if you still want help with this but I recommend you to do this:

 

[math]\int cos^4xdx = \int cos^2x \cdot cos^2x[/math]

 

You know that [math]cos^2x = 1- sin^2x[/math]

 

So you have:

 

[math]\int (1-sin^2x) + \int (1-sin^2x) [/math], then let [math]u=sinx[/math].

 

Hope this helps you abit.

[hotcommodity]

I appreciate the help but I'm a bit lost on what Evo posted. I thought you could only add integrals when something was being added in what was being integrated, rather than multiplied. If you can do it that way I wouldn't be aware of it, I'm only in cal II. Another thing I'm confused on is if you let u=sin(x) then du=cos(x) dx, and there isn't any cos(x) dx to substitute for. Anyway the assignment due tomorrow morning so if anyone wants to work out the problems and test their skills rather than doing the work for me, feel free to do so. I doubt my professor will go over it in class and I'm interested in how to work the problems.

[EvoN1020v]

I appreciate the help but I'm a bit lost on what Evo posted. I thought you could only add integrals when something was being added in what was being integrated, rather than multiplied. If you can do it that way I wouldn't be aware of it, I'm only in cal II. Another thing I'm confused on is if you let u=sin(x) then du=cos(x) dx, and there isn't any cos(x) dx to substitute for. Anyway the assignment due tomorrow morning so if anyone wants to work out the problems and test their skills rather than doing the work for me, feel free to do so. I doubt my professor will go over it in class and I'm interested in how to work the problems.

 

 

I must apologize to you, but I'm not so good in Calculus either, as I'm only in my 1st year in university.

 

But I KNOW this: You need to know how to use chain rule to be able to integrate the powers. Let use the example of [math]\int cos^3x dx[/math], ok?

 

[math]\int cos^3x dx = \int cos^2x \cdot cosx [/math]

 

You know that [math]cos^2x+sin^2x = 1[/math]

 

So you have:

[math]\int (1-sin^2x)cosx[/math]

 

Then you use the Chain Rule, so you let [math]u=sinx[/math] and [math]du=cosxdx[/math].

 

[math] \int (1-u^2)du[/math]

 

Use the Integration Rule, you get:

[math]u - \frac{1}{3}u^3 + C[/math]

 

Replace the u and you have the answer.

[math]sinx - \frac{1}{3}sin^3x + C[/math]

 

 

For your example of [math]cos^4x[/math], you would have [math] \int cos^2x \cdot cos^2x[/math]. Then you do the rest and use [math]u=sinx[/math].

 

Hope this helps!!

[hotcommodity]

It's ok, I really do appreciate the help. I get your example, and u=sin(x) works out great for something like

[math]

\int cos^3x dx

[/math]

 

but when it's [math]\int cos^4x dx[/math] , it's the extra cos(x) that keeps me from letting u=sin(x) because I would have du twice in the problem as du would equal cos(x) and I would have integral (1-sin^2x)cos^2x dx . I wish it was an easy fit like that haha. But I'll see if I can drag the answers out of my professor tomorrow and I'll post them in case anyone is interested. Thanks for everyones posts tho'.

[Klaynos]

I wouldn't use the substitution u, I'd use the identities you're provided just keep subbing in for cos^2x...

[EvoN1020v]

YEAH, I'm interested!! I never thought I am really bad at integrating trigonometric functions.

[EvoN1020v]

After some hours of biting my fingernails trying to solve [math] \int cos^4x[/math], I finally got it. What can I tell you? It's far from your enterprise in what you already know. (You told me you are in high school, right?) In my example of [math]cos^3x[/math], it worked perfectly, because it has an odd power, so you can have a factor of one function. With even power, it's a different story. You have to use the half angle trigonometric identity.

 

[math]\int cos^4x dx = \int (cos^2x)^2[/math]

 

The half angle trig identity for [math]cos^2x[/math] is: [math]\frac{1}{2}(1+cos2x)[/math].

 

You input it in the above equation:

[math] \int (\frac{1}{2}(1+cos2x)^2)^2[/math]

 

It will expands to:

[math]\frac{1}{4} \int (1+ 2cos2x + cos^22x)dx[/math]

 

You should notice that there is another even power for cosx. ([math]cos^22x[/math]) so you have to use the half angle identity again.

 

[math]cos^22x = \frac{1}{2}(1+cos4x)[/math]

 

Input it in the equation:

=[math]\frac{1}{4} \int (1+ 2cos2x + \frac{1}{2}(1+cos^24x))[/math]

 

[math]\frac{1}{4} \int (\frac{3}{2} + 2cos2x + \frac{1}{2}cos4x)dx[/math]

 

Now you can use the Power Rule for Integration:

=[math]\frac{1}{4}(\frac{3}{2}x + sin2x + \frac{1}{8}sin4x)dx[/math]

 

So you have the answer of:

[math]\frac{3}{8}x + \frac{1}{4}sin2x + \frac{1}{32}sin4x + C[/math]

 

You can use the integration checkup at http://integrals.wolfram.com/index.jsp

 

AND IT MATCHES!!

Let me know if you don't understand any of my steps in the calculations. I might have skipped a step.

 

This was definitely out of my enterprise too.. I think I will learn more about this next semester.

 

Anyways, let me know if you have any further questions.

[hotcommodity]

Haha, that's awesome. I like that online integrator too. I wouldn't have thought to use [math]cos^22x = \frac{1}{2}(1+cos4x)[/math] .Great work!

 

Ohhh edit: I'm not in high school, I'm in college btw.

[Klaynos]

After some hours of biting my fingernails trying to solve [math] \int cos^4x[/math], I finally got it. What can I tell you? It's far from your enterprise in what you already know. (You told me you are in high school, right?) In my example of [math]cos^3x[/math], it worked perfectly, because it has an odd power, so you can have a factor of one function. With even power, it's a different story. You have to use the half angle trigonometric identity.

 

[math]\int cos^4x dx = \int (cos^2x)^2[/math]

 

The half angle trig identity for [math]cos^2x[/math] is: [math]\frac{1}{2}(1+cos2x)[/math].

 

You input it in the above equation:

[math] \int (\frac{1}{2}(1+cos2x)^2)^2[/math]

 

It will expands to:

[math]\frac{1}{4} \int (1+ 2cos2x + cos^22x)dx[/math]

 

You should notice that there is another even power for cosx. ([math]cos^22x[/math]) so you have to use the half angle identity again.

 

[math]cos^22x = \frac{1}{2}(1+cos4x)[/math]

 

Input it in the equation:

=[math]\frac{1}{4} \int (1+ 2cos2x + \frac{1}{2}(1+cos^24x))[/math]

 

[math]\frac{1}{4} \int (\frac{3}{2} + 2cos2x + \frac{1}{2}cos4x)dx[/math]

 

Now you can use the Power Rule for Integration:

=[math]\frac{1}{4}(\frac{3}{2}x + sin2x + \frac{1}{8}sin4x)dx[/math]

 

So you have the answer of:

[math]\frac{3}{8}x + \frac{1}{4}sin2x + \frac{1}{32}sin4x + C[/math]

 

You can use the integration checkup at http://integrals.wolfram.com/index.jsp

 

AND IT MATCHES!!

Let me know if you don't understand any of my steps in the calculations. I might have skipped a step.

 

This was definitely out of my enterprise too.. I think I will learn more about this next semester.

 

Anyways, let me know if you have any further questions.

 

Exactly how I would have done it Sorry I didn't explain more clearly I don't have much free time atm