moodydog

02-02-12, 11:11

integrate ((4x-1) / (2x))

Hmm... the boundaries are between 8 and 2

If we split this up into (2x - 1/2x) and differentiate I get 2x - 1/2 ln(2x).

8

[2x -1/2 ln(2x)]

2

[16 - 1/2 ln(16)] - [4 - 1/2 ln(4)]

16 - 1/2 ln(16) - 4 + 1/2 ln(4)

12 + 1/2 ln(1/4)

The answer book gives 12 ln2. My answer is the negative of this. Also wolfram gives:

http://i40.************/2cervk8.gif

What has it done?

_____________________________________

sorry new question, the book I am using is actually swimming with errors int he answers.

If you have sin2x it equals out to 2sinxcosx

But if you have sin(x/2) it equals out to 2sin(x/4)cos(x/4)??? Why?

Thinking about it

Sin(x+x) = sinxcosx + sinxcosx hence 2sinxcosx

but then

Sin(x/2+x/2) should I imagine = 2sin(x/2)cos(x/2) and not 2sin(x/4)cos(x/4). Why is the two doubled? Same with anying else, sinx/6 = 2sin(x/12)cos(x/12)

Why is it doubled?

Secondly

Solving tan2x + tanx =0

I get X as 71.56, 251.56, 0 and 180

the way I did that:

2tanx/(1-tan^x) + tanx = 0

2tanx + tanx(1-tan^2x) = 0

2tanx + tanx - tan^3x = 0

tan^3x - 3tanx = 0

tan^2x - 3 = 0

tanx(tanx-3) = 0

tanx = 0 or tanx = 3

x = 0 or x= 71.56

Between 0 and 360

I get X as 71.56, 251.56, 0 and 180

I got two of those values right, but apparently tanx does not equal three? What is wrong with my method? Or like I said, the book is filled with errors.

Hey again, further down the page, there was this. :o

It uses the same principles as before, I have tried changing the cosec into (1 + cot) with ^2 theta, and tan into sec -1 and sin into 1 - cos, and then expanded everything out, but just came into a big mess.

and also tried the right hand side, and couldn't get further then sec -1...

Sorry this is a newer question I am stuck on;

The question wants me to rearrange the left hand side so it looks identical to the right, using trig identities.

hey, I am stuck on this question;

Solve to find x

(x-2)^2 = (root x) +2

I know one of the answers is 4, but I don't know how to get it. :(

Hmm... the boundaries are between 8 and 2

If we split this up into (2x - 1/2x) and differentiate I get 2x - 1/2 ln(2x).

8

[2x -1/2 ln(2x)]

2

[16 - 1/2 ln(16)] - [4 - 1/2 ln(4)]

16 - 1/2 ln(16) - 4 + 1/2 ln(4)

12 + 1/2 ln(1/4)

The answer book gives 12 ln2. My answer is the negative of this. Also wolfram gives:

http://i40.************/2cervk8.gif

What has it done?

_____________________________________

sorry new question, the book I am using is actually swimming with errors int he answers.

If you have sin2x it equals out to 2sinxcosx

But if you have sin(x/2) it equals out to 2sin(x/4)cos(x/4)??? Why?

Thinking about it

Sin(x+x) = sinxcosx + sinxcosx hence 2sinxcosx

but then

Sin(x/2+x/2) should I imagine = 2sin(x/2)cos(x/2) and not 2sin(x/4)cos(x/4). Why is the two doubled? Same with anying else, sinx/6 = 2sin(x/12)cos(x/12)

Why is it doubled?

Secondly

Solving tan2x + tanx =0

I get X as 71.56, 251.56, 0 and 180

the way I did that:

2tanx/(1-tan^x) + tanx = 0

2tanx + tanx(1-tan^2x) = 0

2tanx + tanx - tan^3x = 0

tan^3x - 3tanx = 0

tan^2x - 3 = 0

tanx(tanx-3) = 0

tanx = 0 or tanx = 3

x = 0 or x= 71.56

Between 0 and 360

I get X as 71.56, 251.56, 0 and 180

I got two of those values right, but apparently tanx does not equal three? What is wrong with my method? Or like I said, the book is filled with errors.

Hey again, further down the page, there was this. :o

It uses the same principles as before, I have tried changing the cosec into (1 + cot) with ^2 theta, and tan into sec -1 and sin into 1 - cos, and then expanded everything out, but just came into a big mess.

and also tried the right hand side, and couldn't get further then sec -1...

Sorry this is a newer question I am stuck on;

The question wants me to rearrange the left hand side so it looks identical to the right, using trig identities.

hey, I am stuck on this question;

Solve to find x

(x-2)^2 = (root x) +2

I know one of the answers is 4, but I don't know how to get it. :(