HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

[Sy123]

Re: MX2 Integration Marathon

it should be 1/2 instead of 2 in your 4th line

Right you are, thanks.

 

[Makematics]

Re: MX2 Integration Marathon

EDIT:
By using IBP(u=lnx, dv= x^(1/n)dx), I get:

 

[Sy123]

Re: MX2 Integration Marathon

By using IBP, I get something weird like



Probs wrong but worth a try

IBP works, but your second term is incorrect.
When i made the integral I was thinking of a substitution, turns out that it can just be done through normal IBP :s

 

[Makematics]

Re: MX2 Integration Marathon

IBP works, but your second term is incorrect.
When i made the integral I was thinking of a substitution, turns out that it can just be done through normal IBP :s

ah yes, silly me, forgot to integrate haha! is it right now, and does it match with your solution using x=u^n or are the answers different?

 

[Sy123]

Re: MX2 Integration Marathon

ah yes, silly me, forgot to integrate haha! is it right now, and does it match with your solution using x=u^n or are the answers different?

That looks right, a substitution of x=u^n just resolves into a u^n ln u, so you then do by parts then you get the answer.

Most probably the answers are the same.

========

 

[Kipling]

Re: MX2 Integration Marathon

Integrate ln(x) / x^4 from 1 to e

 

[asianese]

Re: MX2 Integration Marathon

IBP

 

[chris_power_96]

Re: MX2 Integration Marathon

for integral of (1+cosx)^.5 make 1+cosx=2cos^(2)x/2, then root it to make it (2)^.5 cos(x/2) integrate to get 4(2)^.5?

 

[asianese]

Re: MX2 Integration Marathon

for integral of (1+cosx)^.5 make 1+cosx=2cos^(2)x/2, then root it to make it (2)^.5 cos(x/2) integrate to get 4(2)^.5?

Once again you need to be mindful of the definition of . It'd be best to draw a graph of and see how many repetitions of a 'base' area you need.

 

[Sy123]

Re: MX2 Integration Marathon

 

[chris_power_96]

Re: MX2 Integration Marathon

Do you times by (1-sinx)^.5 on both sides which gives you (1-sinx)^.5 /cosx, then you use u^2=1-sinx, which ultimately gets you to pi/2?

 

[HeroicPandas]

Re: MX2 Integration Marathon

let t = tan(x/2)

*Skip working out*

 

[chris_power_96]

Re: MX2 Integration Marathon

For your t results wouldn't there be a root (1+t^2) on the top though because 1+sinx would equal (1+t)^2/1+t^2, and the other 1+t^2 (which you used) im assuming is from dx=2dt/1+t^2 and also where is the two from that

 

[Sy123]

Re: MX2 Integration Marathon

Do you times by (1-sinx)^.5 on both sides which gives you (1-sinx)^.5 /cosx, then you use u^2=1-sinx, which ultimately gets you to pi/2?

The cos x is on the denominator, and so you cannot substitute du in easily without creating another square root.

 

[chris_power_96]

Re: MX2 Integration Marathon

The cos x is on the denominator, and so you cannot substitute du in easily without creating another square root.

Yeh but there already is one cosx on the denominator so you put in the dx=-2udu/cosx, meaning the denominator goes to cos^2x, then through the fact that sinx =1-u^2, you see that cosx=root(1-(1-u^2)^2)^.5, meaning cos^2x=1-(1-u^2)^2=2u^2-u^4

Then you get the integral 2u^2/(2u^2-u^4)

 

[Sy123]

Re: MX2 Integration Marathon

Yeh but there already is one cosx on the denominator so you put in the dx=-2udu/cosx, meaning the denominator goes to cos^2x, then through the fact that sinx =1-u^2, you see that cosx=root(1-(1-u^2)^2)^.5, meaning cos^2x=1-(1-u^2)^2=2u^2-u^4

Then you get the integral 2u^2/(2u^2-u^4)

yep I see, well then there you go.

My method was using int f(x) = int f(a-x), then applying double angle formula to get a secant integration

 

[chris_power_96]

Re: MX2 Integration Marathon

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

 

[asianese]

Re: MX2 Integration Marathon

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

 

[chris_power_96]

Re: MX2 Integration Marathon

integrate from 0 to pi/2 (1+sin2x)^.5

 

[HeroicPandas]

Re: MX2 Integration Marathon

integrate from 0 to pi/2 (1+sin2x)^.5