HSC 2015 MX2 Integration Marathon (archive) (4 Viewers)

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

studynerd said:
I think this way is faster.. or maybe i just dont like drawing triangles..

let u^2=1+sin^2.... make cot(x)=cos(x)/sin(x)... the you can simply put in the 2udu... it all falls out after that

wheres the actual substitution in your solution?

studynerd · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
wheres the actual substitution in your solution?

At the start

Just let u^2=1+sinx

VBN2470 · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Evaluate $\int \sqrt{1+\sin(x)}\cot(x)dx$$

$Let $u^2 = 1+\sin(x)$, then \\ $\int \sqrt{1+\sin(x)}\cot(x)dx$ \\ = $\int \frac{2u^2}{u^2-1}du$ \\ = $2u + \ln\frac{u-1}{u+1}+C $ \\ = 2\sqrt{1+\sin(x)} + \ln\left(\frac{\sqrt{1+\sin(x)}-1}{\sqrt{1+\sin(x)}+1}\right)+C$$

Ekman · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
$Let $u^2 = 1+\sin(x)$, then \\ $\int \sqrt{1+\sin(x)}\cot(x)dx$ \\ = $\int \frac{2u^2}{u^2-1}du$ \\ = $2u + \ln\frac{u-1}{u+1}+C $ \\ = 2\sqrt{1+\sin(x)} + \ln\left(\frac{\sqrt{1+\sin(x)}-1}{\sqrt{1+\sin(x)+1}\right)+C$$

Whoops saw the mistake in my working out. I saw it as u rather than u^2

VBN2470 · Apr 3, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION

$Find the exact value of $\int_{0}^{1} \frac{\ln(1+x)}{1+x^2}$$

SpiralFlex · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Hello guys haven't posted in a while so don't know where 2015 are up to in integration but it looks like you can have some substitution techniques already. Here are some that should be doable.

$1. \;\textup{Suppose} \;n \;\textup{is an integer greater than or equal to 1. Compute the following integral}$

$\int x^{\frac{1}{n}}\ln(x)}\;dx$

Editing second integral. 1 sec.

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

tomatoplant said:
How do u know what triangles with what sides to form?

It comes from doing triangle questions i just know... Ekman said that the moment i introduced triangles it became wrong... ?

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
$Let $u^2 = 1+\sin(x)$, then \\ $\int \sqrt{1+\sin(x)}\cot(x)dx$ \\ = $\int \frac{2u^2}{u^2-1}du$ \\ = $2u + \ln\frac{u-1}{u+1}+C $ \\ = 2\sqrt{1+\sin(x)} + \ln\left(\frac{\sqrt{1+\sin(x)}-1}{\sqrt{1+\sin(x)}+1}\right)+C$$

Still dont see my mistake

SpiralFlex · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION

$Find the exact value of $\int_{0}^{1} \frac{\ln(1+x)}{1+x^2}$$

Hint for others, exact value is a dead give away especially with zero at the lower limits.

$\int_{0}^{a}f(x)\;dx = \int_{0}^{a}f(a-x)\;dx$

Combine this with a clever substitution (trig).

SpiralFlex · Apr 3, 2015

Re: MX2 2015 Integration Marathon

tomatoplant said:
$u = lnx --- dv = x^\frac{1}{n} dx \\ du = \frac{dx}{x} --- v = \frac{x^\frac{n+1}{n}}{\frac{n+1}{n}} \\ $ I $ = lnx \frac{x^\frac{n+1}{n}}{\frac{n+1}{n}} - \int{\frac{x^{\frac{2n+1}{n}}}{\frac{n+1}{n}} \\ = ... \\ $ sorry for the dashes in 1st line, dunno how to space with latex $$

My mistake, I was intended to make the question so you can't do it by IBP so you would need to make a substitution to do this question. I didn't check that you could do it by IBP.

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

tomatoplant said:
like how did u know that one side was $\sqrt{-(1-x)}$ ?

i see my mistake ahahaha

Ekman · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION

$Find the exact value of $\int_{0}^{1} \frac{\ln(1+x)}{1+x^2}$$

Bump

integral95 · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION

$Find the exact value of $\int_{0}^{1} \frac{\ln(1+x)}{1+x^2}$$

$$Let$ \ \ \ x = tan\theta \ \ \ \ \Rightarrow \int_{0}^{\frac{\pi}{4}} {\ln(1+tan\theta)} = I \\ \\ \\ = \int_{0}^{\frac{\pi}{4}} {\ln(1+tan(\frac{\pi}{4}-\theta)) = \\ \\ \int_{0}^{\frac{\pi}{4}} \ln\frac{2}{(1+tan\theta)} = \int_{0}^{\frac{\pi}{4}} {\ln(2)-ln(1+tan\theta)} \ \ \ \Rightarrow \\ \\ \\ 2I = \int_{0}^{\frac{\pi}{4}} \ln(2) = ln(2)\frac{\pi}{4} \Rightarrow I = ln(2)\frac{\pi}{8}$

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

I got to $\int_0^{\frac{\pi}{4}}{ln(1+tan\theta)d\theta$

Drsoccerball · Apr 4, 2015

Re: MX2 2015 Integration Marathon

New question!
$\int{\sqrt{x}e^{\sqrt{x}}dx$

swagswagyoloswag · Apr 4, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
New question!
$\int{\sqrt{x}e^{\sqrt{x}}dx$

$let $ u^{2} = x \\ 2udu = dx \\ I = \int{2u^2 e^{u}du \\ $ followed by IBP$

VBN2470 · Apr 4, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
New question!
$\int{\sqrt{x}e^{\sqrt{x}}dx$

$Use the substitution $u = \sqrt{x}$, then apply IBP twice to get \\ $2e^{\sqrt{x}}(x-2\sqrt{x}+2)+C$$

VBN2470 · Apr 4, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

$Evaluate $\int \frac{\sec\theta+\tan\theta}{\csc\theta+\cot\theta}d\theta$$

Drsoccerball · Apr 4, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Evaluate $\int \frac{\sec\theta+\tan\theta}{\csc\theta+\cot\theta}d\theta$$

s=sin
c=cos
$= \frac{s(1+s)}{c(1+c)}$

$= \frac{s}{c(1+c)} + \frac{1-c^2}{c(1+c)}$

$$ first fraction $ let u = cosx$

$$ Second fraction $=\frac{1}{c} -1$

$\int{ \frac{-1}{u^2 +u}} + \int{secx -1}$
Where u =cosx

Rest is smoke and mirrors

nightweaver066 · Apr 4, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Evaluate $\int \frac{\sec\theta+\tan\theta}{\csc\theta+\cot\theta}d\theta$$

I see you're on an integration bee spree with your questions haha

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