Evaluating the integral, correct?

[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 :)

 

[rocomath]

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 :)

It wasn't so much irritating, just disappointing. I just want you to get a real good grade :-] I'm mainly pushing you to check your answer to exericse your differentiating skills. I'm more than happy to help!

 

[Zack88]

hey when completing the square when a square roots is involved, do you square both sides?

 

[rocomath]

Example ...

$$x^2-\sqrt 2 x+4$$

This kind of problem?

 

[Zack88]

more like 1 / [sqrt]9x^2 + 6x -8[/sqrt]

 

[rocomath]

You don't have to worry about squaring the square root. Just work within your square root and perform your operation as you would.

But don't forget that your coefficient of your highest degree must be 1.