HSC 2012-14 MX2 Integration Marathon (archive) (4 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) (4 Viewers)

Sy123

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