MX2 Integration Marathon (1 Viewer)

stupid_girl · Jul 17, 2021

a practice on surd manipulation

$\int_{\frac{1}{2}}^{1}\frac{\sin^{-1}x}{2+\sqrt{1+x}+\sqrt{1-x}}dx=\sqrt{2}-\sqrt{6}+\frac{\left(-2+11\sqrt{2}+2\sqrt{3}-3\sqrt{6}\right)\pi}{12}+\ln\left(20-12\sqrt{2}+10\sqrt{3}-8\sqrt{6}\right)-\frac{\pi^{2}}{9}$

stupid_girl · Jul 25, 2021

a slight variant
$\int_{0}^{\frac{\pi}{4}}\frac{\left(\sin2x\right)\left(\sin^{-1}\left(\cos^{2}x\right)\right)}{2+\sin x+\sqrt{1+\cos^{2}x}}dx=\sqrt{2}-\sqrt{6}+\frac{\left(-2+11\sqrt{2}+2\sqrt{3}-3\sqrt{6}\right)\pi}{12}+\ln\left(20-12\sqrt{2}+10\sqrt{3}-8\sqrt{6}\right)-\frac{\pi^{2}}{9}$

For $0\le x\le\pi$ ,
$\int\frac{\left(\sin2x\right)\left(\sin^{-1}\left(\cos^{2}x\right)\right)}{2+\sin x+\sqrt{1+\cos^{2}x}}dx=\frac{\left(\sin^{-1}\cos^{2}x\right)^{2}}{2}-2\left(\sin^{-1}\cos^{2}x\right)\sin\frac{\sin^{-1}\cos^{2}x}{2}-4\cos\frac{\sin^{-1}\cos^{2}x}{2}-\left(\sin^{-1}\cos^{2}x\right)\tan\frac{\sin^{-1}\cos^{2}x}{4}+4\ln\left(\sec\frac{\sin^{-1}\cos^{2}x}{4}\right)+c$

stupid_girl · Sep 18, 2021

Feel free to share your attempt.
$\int_{\frac{1}{4}}^{\frac{1}{2}}\frac{1+2^{8x-1}\cos x^{x}+\ln x+256^{x}+\left(\cos\frac{16^{x}\pi}{3}\right)\ln\sqrt{ex}}{8+4\cos\frac{16^{x}\pi}{3}+4\cos x^{x}+\cos\left(\frac{16^{x}\pi}{3}+x^{x}\right)+\cos\left(\frac{16^{x}\pi}{3}-x^{x}\right)}dx=\frac{3\sqrt{3}}{16\ln2}$

stupid_girl · Oct 21, 2021

Nav123 · Apr 11, 2022

stupid_girl said:
$\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\sqrt{\cos4x}}dx=\frac{3\pi\ln2}{16}$

$I=\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\sqrt{\cos4x}}dx$
$I=\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)^2\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\left(\sin x-\cos x\right)\sqrt{\cos4x}}dx$
$I=\int_{0}^{\frac{\pi}{8}}\frac{\left(1-\sin{2x}\right)\ln\left(\tan2x\right)}{-\cos{2x}\sqrt{\cos4x}}dx$
Let $t=\tan{2x}$
$I=-\frac{1}{2}\int_{0}^{1}\frac{\left(1-\frac{t}{\sqrt{1+t^2}}\right)\ln{t}}{\frac{1}{\sqrt{1+t^2}}\sqrt{\frac{1-t^2}{1+t^2}}}\cdot\frac{1}{1+t^2}dt$
$I=-\frac{1}{2}\int_{0}^{1}\frac{1}{\sqrt{1-t^2}}\ln{t}-\frac{t}{\sqrt{1-t^4}}\ln{t}dt$
Let $J=\int_{0}^{1}\frac{\ln{t}}{\sqrt{1-t^2}}dt$ and $K=\int_{0}^{1}\frac{t\ln{t}}{\sqrt{1-t^4}}dt$

$J=\int_{0}^{\frac{\pi}{2}}\ln{\sin{\theta}}d\theta$ where $t=\sin{\theta}$
$J=\int_{0}^{\frac{\pi}{2}}\ln{\sin{\left(\frac{\pi}{2}-\theta\right)}}d\theta=\int_{0}^{\frac{\pi}{2}}\ln{\cos{\theta}d\theta$ by 'King Property'
$\therefore 2J=\int_{0}^{\frac{\pi}{2}}\ln{\sin{\theta}}+\ln{\cos{\theta}}d\theta=\int_{0}^{\frac{\pi}{2}}\ln{\left(\frac{\sin{2\theta}}{2}\right)}d\theta$
$2J=\frac{1}{2}\int_{0}^{\pi}\ln{\sin{\theta}}d\theta-\int_{0}^{\frac{\pi}{2}}\ln{2}d\theta$ replacing theta with theta/2
$2J=\int_{0}^{\frac{\pi}{2}}\ln{\sin{\theta}}d\theta-\int_{0}^{\frac{\pi}{2}}\ln{2}d\theta$ by symmetry of the sine graph.
$2J=J-\frac{\pi}{2}\ln{2}$
$J=-\frac{\pi}{2}\ln{2}$

Similarily:
$K=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\ln{\sqrt{\sin{\theta}}}d\theta$ letting $t^2=\sin{\theta}$
$K=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}\ln{\sin{\theta}}d\theta$
$K=-\frac{\pi}{8}\ln{2}$

$\therefore I=-\frac{1}{2}\left(-\frac{\pi}{2}\ln{2}+\frac{\pi}{8}\ln{2}\right)=\frac{3\pi\ln2}{16}$

stupid_girl · Jun 4, 2022

sharky564 · Sep 5, 2022

Canteen · Oct 30, 2022

stupid_girl said:
$\int_{0}^{\pi}\left(3\pi-\cos x-2x\right)\left(\frac{x}{1+\sin x}\right)^{2}dx$

To get this thread going again I thought I'd give one of these a go.

Letting $u = \tfrac{\pi}{2} - x$ :
$\begin{aligned}I &= \int_{0}^{\pi}\left(3\pi-\cos x-2x\right)\left(\frac{x}{1+\sin x}\right)^{2}dx \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(2\pi-\sin u + 2u \right)\left(\frac{\frac{\pi}{2} - u}{1+\cos u}\right)^{2}du \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(2u - \sin u + 2u)(\tfrac{\pi^2}{4} - \pi u + u^2)}{(1+\cos u)^2} du \end{aligned}$

We can eliminate any odd function terms in the expansion of the numerator (note that the denominator is even) as we are integrating over $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ , so they go to 0. We end up with:

$\begin{aligned} I &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\tfrac{\pi^3}{2} + 2\pi u^2 + \pi u \sin u - 2\pi u^2}{(1+\cos u)^2} du \\ &= \frac{\pi^3}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{(1+\cos u)^2} du + \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{u \sin u}{(1+\cos u)^2} du \end{aligned}$

Both integrals are pretty straight forward to evaluate now, yielding a final answer of $\tfrac{2\pi^3}{3} + \pi^2 - 2\pi$ .

chilli 412 · Oct 30, 2022

Canteen said:
To get this thread going again I thought I'd give one of these a go.

Letting $u = \tfrac{\pi}{2} - x$ :
$\begin{aligned}I &= \int_{0}^{\pi}\left(3\pi-\cos x-2x\right)\left(\frac{x}{1+\sin x}\right)^{2}dx \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(2\pi-\sin u + 2u \right)\left(\frac{\frac{\pi}{2} - u}{1+\cos u}\right)^{2}du \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(2u - \sin u + 2u)(\tfrac{\pi^2}{4} - \pi u + u^2)}{(1+\cos u)^2} du \end{aligned}$

We can eliminate any odd function terms in the expansion of the numerator (note that the denominator is even) as we are integrating over $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$ , so they go to 0. We end up with:

$\begin{aligned} I &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\tfrac{\pi^3}{2} + 2\pi u^2 + \pi u \sin u - 2\pi u^2}{(1+\cos u)^2} du \\ &= \frac{\pi^3}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{(1+\cos u)^2} du + \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{u \sin u}{(1+\cos u)^2} du \end{aligned}$

Both integrals are pretty straight forward to evaluate now, yielding a final answer of $\tfrac{2\pi^3}{3} + \pi^2 - 2\pi$ .

what was your thought process behind that u-sub?

MX2 Integration Marathon (1 Viewer)

stupid_girl

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stupid_girl

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stupid_girl

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Nav123

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stupid_girl

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sharky564

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Canteen

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chilli 412

oo la la

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