Evaluating the integral, correct?

[Zack88]

now I am lost since its just uv do I just put in uv from the last u and v I got?

 

[rocomath]

$$I=\frac{1}{2}e^{-x}\sin{2x}+\frac{1}{2}\int e^{-x}\sin{2x}dx$$

$$u=e^{-x}$$
$$du=-e^{-x}dx$$

$$dV=\sin{2x}dx$$
$$V=-\frac{1}{2}\cos{2x}$$

$$I=\frac{1}{2}e^{-x}\sin{2x}+\frac{1}{2}\left(-\frac{1}{2}e^{-x}\cos{2x}-\frac{1}{2}\int e^{-x}\cos{2x}dx\right)$$

$$I=\frac{1}{2}e^{-x}\sin{2x}-\frac{1}{4}e^{-x}\cos{2x}-\frac{1}{4}I$$

$$\left(1+\frac{1}{4}\right)I=\frac{1}{2}e^{-x}\sin{2x}-\frac{1}{4}e^{-x}\cos{2x}$$

Now just divide by the constant in front of I and it's solved!

 

[HallsofIvy]

You should not have to "finally take the integral" at all! Please write out exactly what you have after the second integration by parts.

 

[Zack88]

1/2e^-x sin 2x + 1/2(1/2e^-x cos 2x + [integral] 1/2e^-x cos 2x)

 

[rocomath]

Did you not read this post https://www.physicsforums.com/showpost.php?p=1577022&postcount=37

It's practically the answer, you just need to divide the constant :-\

 

[Zack88]

lol yes i did rocophysics, and before i forget I want to think you for all the help I am hoping i can apply this to the other problems i still have to do, I just reposted that because HallsofIvy asked and it never hurts to get more than one opinion.

 

[rocomath]

Oh sorry! Never hurts to get more opinions :-] Exactly why I have like 2 books per subject <3 Amazon and $5 books! If you need anymore help, just post your problems. I'm done doing my hw!

 

[Zack88]

cool i have one question how did you get a 1 + 1/4 in front of I, where did the 1 come from and why isn't 1/4 negative?

 

[rocomath]

cool i have one question how did you get a 1 + 1/4 in front of I, where did the 1 come from and why isn't 1/4 negative?

$$I=\frac{1}{2}e^{-x}\sin{2x}+\frac{1}{2}\int e^{-x}\sin{2x}dx$$

$$u=e^{-x}$$
$$du=-e^{-x}dx$$

$$dV=\sin{2x}dx$$
$$V=-\frac{1}{2}\cos{2x}$$

$$I=\frac{1}{2}e^{-x}\sin{2x}+\frac{1}{2}\left(-\frac{1}{2}e^{-x}\cos{2x}-\frac{1}{2}\int e^{-x}\cos{2x}dx\right)$$

$$I=\frac{1}{2}e^{-x}\sin{2x}-\frac{1}{4}e^{-x}\cos{2x}-\frac{1}{4}I$$

$$\left(1+\frac{1}{4}\right)I=\frac{1}{2}e^{-x}\sin{2x}-\frac{1}{4}e^{-x}\cos{2x}$$

Now just divide by the constant in front of I and it's solved!

It is $$-\frac{1}{4}I$$ I just edited my post to make it a little more clearer.

Also, it's $$1+\frac{1}{4}$$ b/c when I moved the I term from the right to the left, I basically factored out a common term of I rather than finding a common denominator and writing it as one term.

Left hand side $$I + \frac{1}{4}I=\left(1+\frac{1}{4}\right)I$$

 

[Zack88]

so if $$-\frac{1}{4}I$$ was $$\frac{1}{4}I$$ then it would be $$1-\frac{1}{4}I$$ ?

 

[rocomath]

so if $$-\frac{1}{4}I$$ was $$\frac{1}{4}I$$ then it would be $$1-\frac{1}{4}$$ ?

Yep, it's just all Algebra. That's why I started using I, makes it a little easier to see.

$$1\int \cos{2x}dx + \frac{1}{4}\int \cos{2x}dx$$

$$\left(1+\frac{1}{4}\right)\int\cos{2x}dx$$

$$\frac{5}{4}\int\cos{2x}dx$$

 

[Zack88]

ok then for


$$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)$$

how do I go about evaluting since I doesn't reappear. original problem $$\int x^2 \cos mx dx$$it was the first problem we did

 

[rocomath]

ok then for


$$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)$$

how do I go about evaluting since I doesn't reappear. original problem $$\int x^2 \cos mx dx$$it was the first problem we did

Your first problem is the simple, Integrate by Parts once, if that's not enough do it n many times as necessary till you reduce it. The 2nd one is special by the fact that it's "periodic" and your original Integral will reappear after nth time of doing Parts. Basically, just keep doing it till you see a pattern or that it's finally in a simple form that you can Integrate easily.

$$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)$$

This problem is practically solved now, just Integrate your last Integral. You only have cosine mx term, in which m is basically a constant.

$$\frac{1}{m^2}\sin{mx}+C$$

Solved!

 

[Zack88]

$$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)$$

so the -2/m distributes to both -x/m and 1/m so shouldn't $$\frac{1}{m^2}\sin{mx}+C$$ be $$\frac{-2}{m^2}\sin{mx}+C$$

 

[rocomath]

$$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)$$

so the -2/m distributes to both -x/m and 1/m so shouldn't $$\frac{1}{m^2}\sin{mx}+C$$ be $$\frac{-2}{m^2}\sin{mx}+C$$

Yes. Sorry I tend to group things, makes it easier to read and to keep track of things

 

[Zack88]

ok i have another but this time i did a lot and hope I am on the right track

[integral] arctan (1/x)

u = arctan (1/x)
du = -1/((x^2) +1)

dv = dx
v = x

= x arctan(1/x) + [integral] x/((x^2) +1)

u = x^2 +1
du = 2x dx

= x arctan(1/x) + 1/2 [integral] du/u

du/u = ln |u|

= xarctan(1/x) + 1/2 ln |x^2 +1| + c

 

[rocomath]

It's correct :-] Only thing you did wrong was that you forgot dx on your 1st and 2nd Integrals! If you wanted, you can even check your answer by taking it's derivative.

 

[Zack88]

lol the only reason i could do that one is because i was able to follow an example that was close to it, i have another one the weird thing is i got the answer have right

[integral] sec^2 x tan x dx

[integral] (tan^2 x +1) tan x dx

u = tan x
du = sec^2 x dx

[integral] (u^2 +1)u

[integral] u^3 + u

u^4 / 4 + u^2 / 2

my final answer: tan^4 x / 4 + tan^2 x / 2 + c

the real answer: tan^2 x / 2 + c

what did i do wrong?

 

[rocomath]

You need to fix your typographical errors, unless you're already doing so ...

 

[Zack88]

there. i believe

 

[rocomath]

None of those answers are correct, the answer in the book is wrong.

$$\int \sec^{2}x\tan x dx$$

$$\int \sec x \sec x \tan x dx$$

$$u=\sec x dx$$
$$du=\sec x \tan x dx$$

I'm sure you can take it from here.

 

[Zack88]

i get

sec^2 x / 2 + c or 1/2 sec^2 + c

 

[rocomath]

What is the Integral of ...

$$\int \cos^{3}x\sin x dx$$

without working it out on paper?

 

[Zack88]

idk w/o working it out or trying to work it out i know you'll want to get [integral] cos^2 x cos x sinx

[integral] (1 - sin^2) cos x sin x

 

[rocomath]

idk w/o working it out or trying to work it out

Think "chain rule" in reverse. Being able to recognize these types of Integrals will save you a hell of a lot of time in the future.

 

[rocomath]

idk w/o working it out or trying to work it out i know you'll want to get [integral] cos^2 x cos x sinx

Nope!

 

[rocomath]

What is the derivative of ...

$$\frac{d}{dx}(\cos^{4}x)$$

?

 

[Zack88]

-4sin^3?

 

[rocomath]

4cos^3?

You forgot the chain rule!

Review your differentiating methods, it'll pay off in being able to recognize some Integrals.

 

[Zack88]

-4sin^3?

 

[rocomath]

-4sin^3?

You forgot the chain rule!

Review your differentiating methods, it'll pay off in being able to recognize some Integrals.

 

[Zack88]

-4sin x cos^3 x

 

[rocomath]

-4sin x cos^3 x

Yes!

$$-4\cos^{3}x\sin x$$

Now go back to that Integral I asked you about. It's just the chain-rule in "reverse".

 

[Zack88]

$$\int \cos^{3}x\sin x dx$$

- cos^4 x / 4

 

[rocomath]

-4sin^3?

$$\int \cos^{3}x\sin x dx$$

- cos^4 x / 4

Yep. Now look at this problem, but don't pay attention the first part, the 2nd step is what I'm mainly talking about. I spent like an hour evaluating it through Parts a couple times, till I got tired and asked for helped and look how simple it was ...

http://alt1.mathlinks.ro/Forum/latexrender/pictures/7/0/6/70647ef6282a942aac0b3d590d940f1bd06e16d3.gif