HSC 2012-14 MX2 Integration Marathon (archive) (3 Viewers)

Sy123 · Nov 22, 2014

Re: MX2 Integration Marathon

FrankXie said:
$$ At $ x=0, \ln x $ has no definition, so the lower limit $ x=0 $ is understood in the sense of $ x\rightarrow0^+$

$$ Due to the boundedness of sine and cosine function, we have $ \lim_{x\rightarrow0^+}x\sin\ln x=\lim_{x\rightarrow0^+}x\cos\ln x=0$

Now applying integration by parts twice,

$I=\int_0^1\sin\ln xdx=x\sin\ln x|_0^1-\int_0^1x(\cos\ln x)\frac1xdx$

$=0-0-\int_0^1\cos\ln x dx=-x\cos\ln x|_0^1+\int_0^1x(-\sin\ln x)\frac1xdx$

$=-1-0-\int_0^1\sin\ln xdx=-1-I$

$$ Therefore, $ I=-\frac12$

I would of substituted x = e^u, but pretty much the same length (still have to IBP twice)

FrankXie · Nov 22, 2014

Re: MX2 Integration Marathon

yeah this way there is no singularity at x=0

Sy123 · Nov 22, 2014

Re: MX2 Integration Marathon

$\\ $Find$ \ \int_0^{\frac{\pi}{2}} \frac{\sin ((2n+1)x)}{\sin x} \ $d$x \ $where$ \ n \in \mathbb{N}$

FrankXie · Nov 23, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\\ $Find$ \ \int_0^{\frac{\pi}{2}} \frac{\sin ((2n+1)x)}{\sin x} \ $d$x \ $where$ \ n \in \mathbb{N}$

$$ Let $ I_n=\int_0^{\frac{\pi}{2}} \frac{\sin ((2n+1)x)}{\sin x}dx$

$$ Then $ I_n-I_{n-1}=\int_0^{\frac{\pi}{2}} \frac{\sin ((2n+1)x)-\sin ((2n-1)x)}{\sin x}dx$

$=\int_0^{\frac{\pi}{2}}\frac{2\cos((2n)x)\sin x}{\sin x}dx$

$=\left[\frac{\sin((2n)x)}{n}\right]_0^{\frac{\pi}{2}}=0$

$$ Therefore, for all $ n\in \mathcal{N}, I_n=I_1=I_{0}=\frac{\pi}{2}$

Sy123 · Nov 23, 2014

Re: MX2 Integration Marathon

$\\ $Letting$ \ I_{n} = \int_0^{\frac{\pi}{2}} \frac{\sin((2n)x)}{\sin x} \ $d$x \ $where$ \ n \in \mathbb{N}$

$\\ $Find in series form$ \ J_n = \int_0^{\frac{\pi}{4}} \tan^{2n}x \ $d$x \ $where$ \ n \in \mathbb{N}$

$\\ $Hence prove$ \ \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4}$

$\\ $Using the previous result, find$ \ \lim_{n \to \infty} (I_{n+1} - I_{n-1})$

FrankXie · Nov 24, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\\ $Letting$ \ I_{n} = \int_0^{\frac{\pi}{2}} \frac{\sin((2n)x)}{\sin x} \ $d$x \ $where$ \ n \in \mathbb{N}$

$\\ $Find in series form$ \ J_n = \int_0^{\frac{\pi}{4}} \tan^{2n}x \ $d$x \ $where$ \ n \in \mathbb{N}$

$\\ $Hence prove$ \ \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4}$

$\\ $Using the previous result, find$ \ \lim_{n \to \infty} (I_{n+1} - I_{n-1})$

$J_{n+1}+J_n=\int_0^{\frac{\pi}{4}} \tan^{2n}x\sec^2xdx=\int_0^{\frac{\pi}{4}} \tan^{2n}xd\tan x=\frac{1}{2n+1}$(1)$$

$$ so $ \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}=(J_1+J_0)-(J_2+J_1)+(J_3+J_2)-\cdots=J_0\pm J_\infty=\frac{\pi}{4}$

$because $ J_0=\frac{\pi}{4} $ and $ J_\infty=0, $ which follows from equation (1)$

For the last part, I know the limit is zero and I can't see why I need to use previous result. Was there a typo somewhere, or am I mistaken?

FrankXie · Dec 1, 2014

Re: MX2 Integration Marathon

Hey guys we can not rest for so many days. Come warm up with this one:

$$ Evaluate $ \int_0^1\frac{\tan^{-1}x}{1+x}dx.$

integral95 · Dec 1, 2014

Re: MX2 Integration Marathon

FrankXie said:
Hey guys we can not rest for so many days. Come warm up with this one:

$$ Evaluate $ \int_0^1\frac{\tan^{-1}x}{1+x}dx.$

$$using by parts with$ \ \ u = tan^{-1}x \ \ dv = \frac{1}{1+x} \\ \\ \Rightarrow ln(2)\frac{\pi}{4} - \int_{0}^{1}\frac{ln(1+x)}{1+x^2}\ dx \\ \\ $Let I$ \ = \int_{0}^{1}\frac{ln(1+x)}{1+x^2}\ dx \ \ \ \ \ \ u = tan^{-1}x \\ \\ \Rightarrow \int_{0}^{\frac{\pi}{4}}ln(1+tan(u))\ du = \\ \\ \int_{0}^{\frac{\pi}{4}}ln(1+tan(\frac{\pi}{4}-u)) = \int_{0}^{\frac{\pi}{4}}ln(\frac{2}{1+tanu}) = \\ \\ \int_{0}^{\frac{\pi}{4}}ln(2) - ln(1+tanu) \\ \\ \therfore 2I = \int_{0}^{\frac{\pi}{4}}ln(2) = ln(2)\frac{\pi}{4}\Rightarrow I = ln(2)\frac{\pi}{8} \\ \\ \therefore \ \int_0^1\frac{\tan^{-1}x}{1+x} = ln(2)\frac{\pi}{4} - ln(2)\frac{\pi}{8} \\ \\ = ln(2)\frac{\pi}{8}$

FrankXie · Dec 1, 2014

Re: MX2 Integration Marathon

integral95 said:
$$using by parts with$ \ \ u = tan^{-1}x \ \ dv = \frac{1}{1+x} \\ \\ \Rightarrow ln(2)\frac{\pi}{4} - \int_{0}^{1}\frac{ln(1+x)}{1+x^2}\ dx \\ \\ $Let I$ \ = \int_{0}^{1}\frac{ln(1+x)}{1+x^2}\ dx \ \ \ \ \ \ u = tan^{-1}x \\ \\ \Rightarrow \int_{0}^{\frac{\pi}{4}}ln(1+tan(u))\ du = \\ \\ \int_{0}^{\frac{\pi}{4}}ln(1+tan(\frac{\pi}{4}-u)) = \int_{0}^{\frac{\pi}{4}}ln(\frac{2}{1+tanu}) = \\ \\ \int_{0}^{\frac{\pi}{4}}ln(2) - ln(1+tanu) \\ \\ \therfore 2I = \int_{0}^{\frac{\pi}{4}}ln(2) = ln(2)\frac{\pi}{4}\Rightarrow I = ln(2)\frac{\pi}{8} \\ \\ \therefore \ \int_0^1\frac{\tan^{-1}x}{1+x} = ln(2)\frac{\pi}{4} - ln(2)\frac{\pi}{8} \\ \\ = ln(2)\frac{\pi}{8}$

well done

Sy123 · Dec 5, 2014

Re: MX2 Integration Marathon

$\\ \int_0^{\pi /2} \frac{1}{1+ \tan^4 x} \ dx$

HSC 2012-14 MX2 Integration Marathon (archive) (3 Viewers)

Sy123

This too shall pass

FrankXie

Active Member

Sy123

This too shall pass

FrankXie

Active Member

Sy123

This too shall pass

FrankXie

Active Member

FrankXie

Active Member

integral95

Well-Known Member

FrankXie

Active Member

Sy123

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 3)