Evaluating the integral, correct?

[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

 

[Zack88]

I see how you got to step 3 but not to 4

 

[rocomath]

I see how you got to step 3 but not to 4

Oh, don't worry about that. I only wanted to emphasize the chain-rule. Step 3 to 4 would take me forever to type, lol.

I simplified a lot to save myself typing-time. I basically used a trig identity.

 

[Zack88]

lol i was looking at it and then i tilted my head and was like yep i don't know how that happened

 

[Zack88]

woo now i know 6 out of the 18 problems i have to do are correct thanks to you

 

[rocomath]

woo now i know 6 out of the 18 problems i have to do are correct thanks to you

I'm getting sleepy, if you want to do a couple more better start asking :-]

 

[Zack88]

me too me too

[integral] sec^6 x dx

should i do

[integral] (sec^2)^3

or

start doing parts

u= sec^2 x dv = sec^4 x

 

[rocomath]

\int\sec^6 xdx

\int\sec^4 x \sec^2 x dx

\int(\sec^2 x)^2 \sec^2 xdx

When you trig identities raised to powers, break it up till you see something.

 

[Zack88]

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

u = tan x
du = sec^2 x

[integral] (u^2 =1)^2 du

[integral]u^4 + 2u^2 + 1 du

tan^5 x / 5 + 2tan^3 x / 3 + tan x + c

 

[rocomath]

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

u = tan x
du = sec^2 x

[integral] (u^2 =1)^2 du

[integral]u^4 + 2u^2 + 1 du

tan^5 x / 5 + 2tan^3 x / 3 + tan x + c

Good!

 

[Zack88]

woot lol

 

[Zack88]

im goin to let my brain cool down for the night, t2ul and have a good night.

 

[rocomath]

woot lol

Alright, I'm going to sleep! You still have Sunday and Monday to finish all 18 problems!

 

[Zack88]

woo I am back, v.v

[integral] sin^5 x cos^3 x

i was wanting to know if i should break it up like

[integral] sin^4 x sin x cos^2 x cos x

 

[rocomath]

Look in your book, there should be a suggestion on how to tackle even/odd powers of sines and cosines.

Hint: Leave sin^5 x alone, mess around with cos^3 x

 

[Zack88]

[integral] sin^5 x cos^2 x cos x
[integral] sin^5 x (1- sin^2x) cos x

u= sin
du= cos

[integral] u^5(1- u^2) du
[integral] u^5 - u^7 du

1/6 sin^6 x - 1/8 sin^8 x + c

 

[rocomath]

[integral] sin^5 x cos^2 x cos x
sin^5 x (1- sin^2x) cos x
u= sin
du= cos

[integral] u^5(1- u^2) du
u^5 - u^7 du

1/6 sin^6 x - 1/8 sin^8 x + c

Good!

 

[Zack88]

woo, for

[integral] x cos^2 x dx

do i want to use

cos^2 x = (1 + cos^2 x) / 2, then parts

or

[integral] x (1-sin^2x), then parts?

 

[rocomath]

woo, for

[integral] x cos^2 x dx

do i want to use

cos^2 x = (1 + cos^2 x) / 2, then parts

or

[integral] x (1-sin^2x), then parts?

Try a method! Come test day, you got to just go at it :-] You've handled harder problems than this, I'm sure you can do this with ease.

 

[Zack88]

sorry had a fire drill and then went to go eat

im goin to do the (cos^2 x)/ 2 and then the chain rule

 

[Zack88]

[integral] x cox^2 x dx

u= x
du = dx

dv= cos^2
v= (sin(2x)) / 4 + x/2

x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du

x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c

 

[Zack88]

i know the answer is close to that except for the extra x/2

 

[rocomath]

Your identity is wrong

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

 

[Zack88]

thank you must have wrote it down wrong on paper, but does that matter since i didnt use that identity?

 

[rocomath]

thank you must have wrote it down wrong on paper, but does that matter since i didnt use that identity?

What is this supposed to be ...

cos^2 x = (1 + cos^2 x) / 2

?

That's not an identity.

 

[Zack88]

[integral] x cox^2 x dx

u= x
du = dx

dv= cos^2
v= (sin(2x)) / 4 + x/2

x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du

x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c

i know the answer is

x (sin(2x)) / 4 + cos(2x) / 8 + x^2 / 4 + c

but I got an extra x/2

 

[rocomath]

[integral] x cox^2 x dx

u= x
du = dx

dv= cos^2
v= (sin(2x)) / 4 + x/2

x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du

x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c

i know the answer is

x (sin(2x)) / 4 + cos(2x) / 8 + x^2 / 4 + c

but I got an extra x/2

No. Use an identity on cos^2 x. You must reduce the power to 1 before you can integrate it.

 

[Zack88]

ok

[integral] x cox^2 x dx

turns into

[integral] x/2(1 + cos2x) dx

 

[Zack88]

[integral] x/2 (1 + cos2x)
[chain rule]
[integral] x/2 (-2sin2x) + (1 + cos2x)(1/2)
= x^2 /4 (cos2x) + (((sin(2x) / 2) + x)) (x/2)
= (x^2 cos 2x) / 4 + x(sin2x / 2 + x/2)

 

[rocomath]

[integral] x/2 (1 + cos2x)
[chain rule]
[integral] x/2 (-2sin2x) + (1 + cos2x)(1/2)
= x^2 /4 (cos2x) + (((sin(2x) / 2) + x)) (x/2)
= (x^2 cos 2x) / 4 + x(sin2x / 2 + x/2)

Check it by taking the derivative, if I have to tell you again I'm not helping anymore.

 

[Zack88]

sorry if i was irritating you it has been a year since I've done any math and I am jumping into calc 2, but i finally got it and I have done most of the rest, just three left which I am working on now, so thank you for all your help and putting up with me :)