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**Natkirky**- Replies: 10

Hi,

I am doing first year Uni maths. I was wondering if someone could help me with the following question:

∫tan-¹x/1 + x²

I have the solution although I don't understand how to get to it. I would be very grateful if someone could go through all the steps with me.

Thank you

Hi Bob,

That's great I keep getting mixed up with the sec and tan ones. I see what you are doing it's a big help thanks very much.

Hi Bob,

Thanks for your help can I double check you mean u = tan²x and not tanx as I was sure the derivative of tanx was sec²x.

By using the trig identities I managed to get du = sec²x dx which is du = (1 + u²) dx. It was at this point I started to struggle i was not sure where to go from here.

Thanks again for your help

**Natkirky**- Replies: 5

Hi, I am just finishing my first year at Uni.

I was wondering if someone could help me with this question?

∫Tan^4x dx.

I Have a solution for this however, I don't really understand where the values are coming from and I also done a search online and got completely different answers.

We were given this before we done partial fractions and integration by parts so i'm pretty sure the solution should not have anything like that in it.

The solution I have from Uni is 1/6 tan^6x + 1/4 tan^4x + c. By taking u=tanx. Now the answers i have been finding online are 1/3 tan^3x - tanx + x + c, this solution makes sense to me and its close to what i got first time I am just worried it's very different from the solution I have from Uni.

I would be very grateful for any help.

Thank you

**Natkirky**- Replies: 2

Hi,

I am doing first year uni maths and i was wondering if someone could please help with the following question?

By multiplying together known Maclaurin Series obtain the expansion of (cos2x)log(1+x²) up to term x^6.

I have found co2x to be 1- 2x² + 2/3x^4 - 4/45x^6 + ...

and log(1+x²) to be tan-¹x = x - 1/3x³ + 1/5x^5 +...

By multiplying these values together I am not getting the correct answer and I don't have a clue where I'm going wrong. The solution I have is x² - 5/2x^4 + 2x^6.

I would be grateful for any help or corrections if I am doing this completely wrong.

Thank you.

Hi Bob,

Thanks a lot for your help I have a test on vectors tomorrow and that question was bothering me

**Natkirky**- Replies: 3

Hi,

I am doing 1st year university algebra. I was wondering if anyone could tell me if i have done this question correctly?

Q. Let u and v be perpendicular vectors of lengths 1 and 2 respectively. Evaluate (3u+v) (u-2v).

By expanding the bracket I get

3(u.u)-5(u.v)-2(v.v)

Since u is a unit vector u.u=1 and as they are perpendicular u.v=0

Hence 3(1) - 5(0) - 8 = -5.

Is this correct i am unsure of what to do with v.v as v is of length 2 therefore not a unit vector so I just took the square of 2.

I would be grateful for any help with this.

Thanks

Hi gAr,

Thanks for the help that was one of the answers i had.

Thanks a lot

Natkirky

**Natkirky**- Replies: 3

Hi i was wondering if anyone could help me with the following question

∫sin(√ x)/√ x

I have a few ways which i could do this i'm just not sure what is the correct answer.

Thank you

Natkirky

Hi Mathsyperson,

Thanks for the step got that to work out now. I always use the denominator to multiply by did not even think to use this.

Thanks again

Natkirky

**Natkirky**- Replies: 3

Hi,

I was wondering if someone could help with this question:

Show: 1 - cosθ/sinθ = sinθ/1 + cosθ

I can get all other variations of this to work and i can only get this to work the wrong way around ie 1 + cosθ/sinθ, so I know it's something really stupid I am missing.

I would be grateful for any help or tips

Thanks Natkirky

Hi Bobbym,

Thanks a lot i missed the difference of squares in the first bracket as soon as i saw that i figured the rest out.

Thanks for the hints i did know these identities at one point hopefully it will all come back to me.

Thanks

Natalie

**Natkirky**- Replies: 3

Hi Can i please have help simplifying this equation:

(x² - y²)² + 4xy(x+y)²

I am in 1 year at uni however, it's been 4 years since i done maths and i was never great at factorising.

I understand that i am looking for a common factor the solution i have is (x+y)^4 I have tried everything i can think of and can't get this.

Thank you

Hi Bobbym,

That's great thanks for your help.

Natkirky:D

**Natkirky**- Replies: 4

Hi,

Can someone help me with this complex equation

1/Z + 1/2+i = 1/1+3i?

The answer is Z=-3 + i

I have tried everything i can think of and can't get this

Thank you

Hi Bob,

That's great thank you?

Natalie

**Natkirky**- Replies: 3

Hi i was wondering if anyone could help me with this question.

Find integers m and n such that 8m + 26n = 2.

I can't make sense of my notes and don't even know how to start.

Thank you

Hi Bob,

Thanks a lot it has taken me ages to get this I really need to start reading the questions properly!!

Hi Bob,

Thanks for your help i knew it would be something silly never thought about squaring the bracket last.

Can i just double check why e^x squared is e^2x and not e^x².

Thanks again for your help.

Hi Bob,

The question just asks to simplify the expression. It's using the rule e^ ln (something) = something. I can do these but I'm stuck on the squared bracket.

The answer is x^6 e^2x. I hope that helps and thank you.

**Natkirky**- Replies: 8

Hi,

Can someone please show me how to work out this equation? [exp (x+3 ln x)]²

I have the answer but no working and I can't get to it.

Its the whole equation being raised to 2 that's getting me, can you square exp? If so how does this work.

I might be missing something really silly I'm not great with logs and keep forgetting all the different formulas.

Thank you

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