MX2 Integration Marathon (2 Viewers)

stupid_girl · Jun 16, 2020

This one should be easier.
$\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$

stupid_girl · Jul 4, 2020

a new one...shouldn't be too hard

Feel free to share your attempt.

Find the area bounded by x-axis and the curve
$y=\frac{\sqrt{64x^{6}-16x^{4}+x^{2}}}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)\left(1+\pi^{x}\right)}$ .

stupid_girl · Jul 4, 2020

stupid_girl said:
This one should be easier.
$\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$

$\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$
$=\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\frac{\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)+\ln\left(\frac{1}{4}+\frac{3\tan\left(\frac{\pi}{10}+\frac{3\pi}{20}-x\right)-1}{1-3\tan^{2}\left(\frac{\pi}{10}+\frac{3\pi}{20}-x\right)}\right)}{2}dx$
$=\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\frac{3}{4}dx$
$=\frac{\pi}{20}\ln3-\frac{\pi}{10}\ln2$

Nav123 · Sep 26, 2020

stupid_girl said:
a new one...shouldn't be too hard
Feel free to share your attempt.

Find the area bounded by x-axis and the curve
$y=\frac{\sqrt{64x^{6}-16x^{4}+x^{2}}}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)\left(1+\pi^{x}\right)}$ .

Not sure if this is still active but I'll give this a go for fun (let me know if I've made any mistakes or there is an easier way). The x-ints are $x= \pm \frac{1}{2\sqrt{2}}$
Using the corrected identity (Thanks to vernburn for showing that there is a factor of a half):
$\int_{-a}^{a} \frac{E}{1+b^O}dx=\frac{1}{2}\int_{-a}^{a} Edx$
where E is an even function and O is an odd function. The integral simplifies:
$I=\int_{-\frac{1}{2\sqrt{2}}}^{\frac{1}{2\sqrt{2}}}\frac{\sqrt{64x^{6}-16x^{4}+x^{2}}}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)\left(1+\pi^{x}\right)}dx$
$=\frac{1}{2}\int_{-\frac{1}{2\sqrt{2}}}^{\frac{1}{2\sqrt{2}}}\frac{\sqrt{64x^{6}-16x^{4}+x^{2}}}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)}dx$
$=\int_{0}^{\frac{1}{2\sqrt{2}}}\frac{\sqrt{x^2(8x^2-1)^2}\left(\sqrt{1-16x^{4}}-\sqrt{12x^{2}-48x^{4}}\right)}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)\left(\sqrt{1-16x^{4}}-\sqrt{12x^{2}-48x^{4}}\right)}dx$

$\text{The denominator simplfies to:} \hspace{3mm} |1-16x^4|-|12x^2-48x^4|=-(1-16x^4)-(12x^2-48x^4)=(1-8x^2)(1-4x^2)$

$I=\int_{0}^{\frac{1}{2\sqrt{2}}}\frac{x\sqrt{1-16x^{4}}}{1-4x^2}dx-2\sqrt{3}\int_{0}^{\frac{1}{2\sqrt{2}}}\frac{x\sqrt{x^2-4x^{4}}}{1-4x^2}dx$
Let $J=\int_{0}^{\frac{1}{2\sqrt{2}}}\frac{x\sqrt{1-16x^{4}}}{1-4x^2}dx$ and $K=\int_{0}^{\frac{1}{2\sqrt{2}}}\frac{x\sqrt{x^2-4x^{4}}}{1-4x^2}dx$
$J=\frac{1}{8}\int_{0}^{\frac{\pi}{6}}\frac{\cos{u}\sqrt{1-\sin^2{u}}}{1-\sin{u}}du$ where $4x^2=\sin{u}$
$J=\frac{1}{8}\int_{0}^{\frac{\pi}{6}}\frac{(1-\sin{u})(1+\sin{u})}{1-\sin{u}}du$
$J=\frac{1}{8}\left[u-\cos{u}\right]_0^\frac{\pi}{6}=\frac{1}{8}{\left(\frac{\pi}{6}-\frac{\sqrt{3}}{2}+1\right)$

$K=\int_{0}^{\frac{1}{2\sqrt{2}}}\frac{x^2}{\sqrt{1-4x^2}}dx$
$K=\frac{1}{8}\int_{0}^{\frac{\pi}{4}}\frac{\sin^2{u}\cos{u}}{\sqrt{1-\sin^2{u}}}du$ where $2x=\sin{u}$
$K=\frac{1}{16}\int_{0}^{\frac{\pi}{4}}(1-\cos{2u}) du=\frac{1}{16}\left[u-\frac{1}{2}\sin{2u}\right]_0^\frac{\pi}{4}=\frac{1}{16}{\left(\frac{\pi}{4}-\frac{1}{2}\right)$
So $I=\frac{1}{8}{\left(\frac{\pi}{6}-\frac{\sqrt{3}}{2}+1\right)-\frac{\sqrt{3}}{8}{\left(\frac{\pi}{4}-\frac{1}{2}\right)=\frac{\pi}{48}-\frac{\sqrt{3}\pi}{32}+\frac{1}{8}$

stupid_girl · Sep 27, 2020

Your approach is correct but unfortunately a factor of 1/2 is missing somewhere.
$\int_{-\frac{1}{2\sqrt{2}}}^{\frac{1}{2\sqrt{2}}}\frac{\sqrt{64x^{6}-16x^{4}+x^{2}}}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)\left(1+\pi^{x}\right)}dx=\frac{\pi}{48}-\frac{\sqrt{3}}{32}\pi+\frac{1}{8}$

stupid_girl · Sep 27, 2020

another one

Feel free to share your attempt.
$\int_{\frac{1}{4}}^{\frac{1}{2}}\sin\left(x^{x}+\ln x\right)\cos\left(x^{x}-\ln x\right)\log_{2}\left(ex\right)dx=\frac{\left(5\ln2-8\right)\sin\left(2\ln2\right)+\left(4-5\ln2\right)\sin\left(4\ln2\right)+\left(10\ln2-6\right)\cos\left(2\ln2\right)+\left(3-10\ln2\right)\cos\left(4\ln2\right)}{100\ln2}$

Nav123 · Sep 27, 2020

stupid_girl said:
Your approach is correct but unfortunately a factor of 1/2 is missing somewhere.
$\int_{-\frac{1}{2\sqrt{2}}}^{\frac{1}{2\sqrt{2}}}\frac{\sqrt{64x^{6}-16x^{4}+x^{2}}}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)\left(1+\pi^{x}\right)}dx=\frac{\pi}{48}-\frac{\sqrt{3}}{32}\pi+\frac{1}{8}$

Not quite sure where I missed the half, but I might not have checked it properly.

vernburn · Sep 27, 2020

Nav123 said:
Not quite sure where I missed the half, but I might not have checked it properly.

I believe the factor of a half comes from the identity you (incorrectly) quoted. The identity should be:
$\int_{-a}^{a}\frac{E}{1+b^O}\;\mathrm{d}x=\frac{1}{2}\int_{-a}^{a}E\;\mathrm{d}x$
Proof:
$I = \int_{-a}^{a}\frac{E}{1+b^O}\;\mathrm{d}x$
$= - \int_{a}^{-a}\frac{E}{1+b^{-O}}\;\mathrm{d}u \ \ \ (\textrm{using the sub u=-x and the properties of odd and even functions})$
$= \int_{-a}^{a}\frac{E}{1+b^{-O}}\;\mathrm{d}x \ \ \ (\textrm{switching bounds and dummy variable})$
$= \int_{-a}^{a}\frac{Eb^O}{1+b^O}\;\mathrm{d}x$
$\therefore 2I = \int_{-a}^{a}\frac{E(1+b^O)}{1+b^O}\;\mathrm{d}x$
$\implies I = \frac{1}{2}\int_{-a}^{a}E\;\mathrm{d}x$

Qeru · Jan 12, 2021

stupid_girl said:
another one

Feel free to share your attempt.
$\int_{\frac{1}{4}}^{\frac{1}{2}}\sin\left(x^{x}+\ln x\right)\cos\left(x^{x}-\ln x\right)\log_{2}\left(ex\right)dx=\frac{\left(5\ln2-8\right)\sin\left(2\ln2\right)+\left(4-5\ln2\right)\sin\left(4\ln2\right)+\left(10\ln2-6\right)\cos\left(2\ln2\right)+\left(3-10\ln2\right)\cos\left(4\ln2\right)}{100\ln2}$

Using the identity:

$\sin\alpha\cos\beta=\frac{1}{2}\left(\sin{(\alpha+\beta)}+sin{(\alpha-\beta)}\right)$

$\int_{\frac{1}{4}}^{\frac{1}{2}}\sin\left(x^{x}+\ln x\right)\cos\left(x^{x}-\ln x\right)\log_{2}\left(ex\right)dx$

$=\frac{1}{2\ln{2}}\int_{\frac{1}{4}}^{\frac{1}{2}}(\sin(2x^x)+\sin(2\ln{x}))(\ln{e}+\ln{x}) dx$

$=\frac{1}{2\ln{2}}\int_{\frac{1}{4}}^{\frac{1}{2}}\sin(2x^x)(1+\ln{x})+\sin(2\ln{x})(1+\ln{x}) dx$

For the second integral, let $t=\ln{x} \implies \frac{dt}{dx}=\frac{1}{x} \implies dx=e^tdt$

$I=\int_{\frac{1}{4}}^{\frac{1}{2}}\sin(2\ln{x})(1+\ln{x}) dx$

$=\int_{\ln\frac{1}{4}}^{\ln\frac{1}{2}}\sin(2t)(1+t)e^t dt$

$\int e^t\sin(2t) dt=\frac{e^t}{5}(\sin2t-2\cos2t)+C$ and $\int e^u\cos(2t) dt=\frac{e^t}{5}(2\sin2t+\cos2t)+C$ which can be found quite easily using two applications of integration by parts . Now using IBP on the main integral: $u=t+1,v'= e^usin(2t)$ hence $u'=1,v=\frac{e^t}{5}(\sin2t-2\cos2t)$

$I=\left[\frac{e^t}{5}(t+1)(\sin2t-2\cos2t)\right]_{ln\frac{1}{4}}^{ln\frac{1}{2}}{-\int_{ln\frac{1}{4}}^{ln\frac{1}{2}}\frac{e^t}{5}(\sin2t-2\cos2t) dt$

$I=\left[\frac{e^t}{5}(t+1)(\sin2t-2\cos2t)-\frac{e^t}{25}(\sin2t-2\cos2t)+\frac{2e^t}{25}(2\sin2t+\cos2t)\right]_{ln\frac{1}{4}}^{ln\frac{1}{2}}$ using the previous results

$I=\left[\frac{e^t}{5}\left(\left(t+\frac{4}{5}\right)(\sin2t-2\cos2t)+\frac{2}{5}(2\sin2t+\cos2t)\right)\right]_{ln\frac{1}{4}}^{ln\frac{1}{2}}$

$I=\left[\frac{e^t}{5}\left(\left(t+\frac{8}{5}\right)\sin2t-\left(2t+\frac{6}{5}\right)\cos2t\right)\right]_{ln\frac{1}{4}}^{ln\frac{1}{2}}$

$I=\left(\frac{1}{10}-\frac{1}{20}\right)\left(\left(\frac{-5\ln{2}+8}{5}\right)\sin{(-2\ln{2})-\left(\frac{-5\ln{4}+8}{5}\right)\sin{(-2\ln{4})}}-\left(\frac{-10\ln{2}+6}{5}\right)\cos{(-2\ln{2}})+\left(\frac{-10\ln{4}+6}{5}\right)\cos{(-2\ln{4})}\right)$

$I=\frac{\left(5\ln2-8\right)\sin\left(2\ln2\right)+\left(8-10\ln2\right)\sin\left(4\ln2\right)+\left(10\ln2-6\right)\cos\left(2\ln2\right)+\left(6-20\ln2\right)\cos\left(4\ln2\right)}{100}$ Using the fact that sine is odd and cosine is even.

Now for the first integral letting $u=x^x$ gives identical bounds $\frac{1}{\sqrt{2}}$ hence $\int_{\frac{1}{4}}^{\frac{1}{2}}\sin(2x^x)(1+\ln{x}) dx=\int_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{\sin(2u)}{u} du=0$

So in total: $\int_{\frac{1}{4}}^{\frac{1}{2}}\sin\left(x^{x}+\ln x\right)\cos\left(x^{x}-\ln x\right)\log_{2}\left(ex\right)dx=\frac{\left(\frac{5\ln2-8}{2}\right)\sin\left(2\ln2\right)+\left(4-5\ln2\right)\sin\left(4\ln2\right)+\left(5\ln2-3\right)\cos\left(2\ln2\right)+\left(3-10\ln2\right)\cos\left(4\ln2\right)}{100\ln{2}}$

Wow that was a long one, unfortunately didn't get it out as I'm off by a bit, can anyone see my error?

stupid_girl · Jan 16, 2021

Qeru said:
$I=\left[\frac{e^t}{5}\left(\left(t+\frac{8}{5}\right)\sin2t-\left(2t+\frac{6}{5}\right)\cos2t\right)\right]_{ln\frac{1}{4}}^{ln\frac{1}{2}}$

$I=\left(\frac{1}{10}-\frac{1}{20}\right)\left(\left(\frac{-5\ln{2}+8}{5}\right)\sin{(-2\ln{2})-\left(\frac{-5\ln{4}+8}{5}\right)\sin{(-2\ln{4})}}-\left(\frac{-10\ln{2}+6}{5}\right)\cos{(-2\ln{2}})+\left(\frac{-10\ln{4}+6}{5}\right)\cos{(-2\ln{4})}\right)$

Wow that was a long one, unfortunately didn't get it out as I'm off by a bit, can anyone see my error?

This is the product of two functions. You cannot substitute and multiply this way.

Qeru · Jan 29, 2021

Since this thread is dead I thought I would post a question: Find: $\int \frac{1}{\sqrt{x\sqrt{x}-x^2}} dx$

Randomstudent123 · Jan 29, 2021

Is this correct?

Qeru · Jan 29, 2021

Randomstudent123 said:
View attachment 30150

Is this correct?

Yes well done guess it was too easy Try:

$\int \frac{1}{\sqrt[4]{x^4+1}} dx$

vernburn · Jan 29, 2021

Qeru said:
Since this thread is dead I thought I would post a question: Find: $\int \frac{1}{\sqrt{x\sqrt{x}-x^2}} dx$

I did this a slightly different way:
$I=\int \frac{1}{\sqrt{x\sqrt{x}-x^2}} dx$
$=\int \frac{1}{\sqrt{x}\sqrt{\sqrt{x}-x}} dx$
$=2\int \frac{1}{2\sqrt{x}\sqrt{\frac14-\left(\sqrt{x}-\frac12\right)^2}} dx$
You will notice this is in the form $\int \frac{f'(x)}{\sqrt{a^2-\left[f(x)\right]^2}}dx=\sin^{-1}\left(\frac{f(x)}{a}\right)+C$ , yielding:
$2\sin^{-1}\left(2\sqrt{x}-1\right)+C$
I thought I had gotten it wrong at first because it differed from @Randomstudent123 's answer but I checked on Desmos and the graphs differ by $\pi$ and so it is indeed correct.

vernburn · Jan 30, 2021

Qeru said:
Yes well done guess it was too easy Try:

$\int \frac{1}{\sqrt[4]{x^4+1}} dx$

This is my first attempt but the answer‘s slightly off. On the line marked $(\ast)$ , if the denominator is $u^4-1$ instead of $1-u^4$ then I think the final answer is correct but I can’t find the missing negative.

notme123 · Jan 30, 2021

vernburn said:
This is my first attempt but the answer‘s slightly off. On the line marked $(\ast)$ , if the denominator is $u^4-1$ instead of $1-u^4$ then I think the final answer is correct but I can’t find the missing negative.

I dont think you converted from theta to x correctly. Shouldn't $(\sqrt{sin(theta)})=\frac{x}{\sqrt[4]{(x^4+1)}}$ , you did the reciprocal I think. Other than that you got it right.

notme123 · Jan 30, 2021

Qeru said:
Yes well done guess it was too easy Try:

$\int \frac{1}{\sqrt[4]{x^4+1}} dx$

Also @Qeru pls post more I was SOOOO close to getting $\int\frac{dx}{\sqrt[4]{x^4+1}$ . I used the right sub but my mind just died when I got to the part right before u-sub. I did the right u-sub as well I just did a dumb mistake with the differentials as its weird to work with $d(Theta)$ and $du$

vernburn · Jan 30, 2021

notme123 said:
I dont think you converted from theta to x correctly. Shouldn't $(\sqrt{sin(theta)})=\frac{x}{\sqrt[4]{(x^4+1)}}$ , you did the reciprocal I think. Other than that you got it right.

Yes, you are correct. I’m so dumb!

Qeru · Jan 30, 2021

Evaluate:
$\int_{0}^{\infty}\frac{\ln{x}}{1+x^2} dx$

vernburn · Jan 30, 2021

Qeru said:
Evaluate:
$\int_{0}^{\infty}\frac{\ln{x}}{1+x^2} dx$

Firstly, use the substitution $x=\frac{1}{u}\implies dx = -\frac{1}{u^2}du$ :
$I = -\int_{\infty}^{0}\frac{\ln{\frac{1}{u}}}{1+\frac{1}{u^2}} \cdot\frac{du}{u^2}$
$=\int_0^{\infty}\frac{-\ln{u}}{u^2+1}du$
$=-I$
Thus,
$2I=0\implies I=0$

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