# MX2 Integration Marathon (1 Viewer)

#### stupid_girl

##### Active Member
$\bg_white \int_{0}^{\frac{\pi}{4}}\left(\sin^{-1}\left(\tan x\right)+\tan^{-1}\left(\sin2x\right)+\tan^{-1}\left(\cos2x\right)\right)dx$

#### Jason Pham

##### New Member
I’m quite stuck with this question, been doing for week. Pretty appreciate your help

#### Attachments

• 74 KB Views: 75

#### Jason Pham

##### New Member
$\bg_white \int_{0}^{\frac{\pi}{4}}\left(\sin^{-1}\left(\tan x\right)+\tan^{-1}\left(\sin2x\right)+\tan^{-1}\left(\cos2x\right)\right)dx$

#### Attachments

• 1.3 MB Views: 61

#### Jason Pham

##### New Member
$\bg_white \int_{0}^{\frac{\pi}{4}}\left(\sin^{-1}\left(\tan x\right)+\tan^{-1}\left(\sin2x\right)+\tan^{-1}\left(\cos2x\right)\right)dx$
Hopefully I can introduce you another approach which is pure substitution and differential integration that I’m developing. However i need to prepare for trials atm!

#### quickoats

##### Well-Known Member
I’m quite stuck with this question, been doing for week. Pretty appreciate your help
Use the fact that for $\bg_white 0 \leq x \leq 1, \; xf(x) \leq f(x)$ with equality holding for all x iff f(x) = 0.

#### quickoats

##### Well-Known Member
Note that $\bg_white \frac{\sin^{-1} x}{\cos^{-1} x} \neq \tan^{-1} x$ so some of your working is incorrect.

#### Jason Pham

##### New Member
Use the fact that for $\bg_white 0 \leq x \leq 1, \; xf(x) \leq f(x)$ with equality holding for all x iff f(x) = 0.
I see I’ll check. Cuz i later figured out it’s Rolle theorem. Thank you, I’ll check them out

#### Jason Pham

##### New Member
Note that $\bg_white \frac{\sin^{-1} x}{\cos^{-1} x} \neq \tan^{-1} x$ so some of your working is incorrect.
I see, yeah i just realized. Thank you, I’ll fix it

#### stupid_girl

##### Active Member
Hopefully I can introduce you another approach which is pure substitution and differential integration that I’m developing. However i need to prepare for trials atm!

##### -insert title here-
$\bg_white \int_{0}^{\frac{\pi}{4}}\left(\sin^{-1}\left(\tan x\right)+\tan^{-1}\left(\sin2x\right)+\tan^{-1}\left(\cos2x\right)\right)dx$
i like how this is just a reskin of the BOS Trials' cancerous arctangent integral

#### stupid_girl

##### Active Member
a practice on surd manipulation
$\bg_white \int_{\frac{1}{2}}^{1}\frac{\sin^{-1}x}{2+\sqrt{1+x}+\sqrt{1-x}}dx=\sqrt{2}-\sqrt{6}+\frac{\left(-2+11\sqrt{2}+2\sqrt{3}-3\sqrt{6}\right)\pi}{12}+\ln\left(20-12\sqrt{2}+10\sqrt{3}-8\sqrt{6}\right)-\frac{\pi^{2}}{9}$

#### stupid_girl

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

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

#### stupid_girl

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

#### stupid_girl

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

#### Nav123

##### Member
$\bg_white \int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\sqrt{\cos4x}}dx=\frac{3\pi\ln2}{16}$
$\bg_white I=\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\sqrt{\cos4x}}dx$
$\bg_white I=\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)^2\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\left(\sin x-\cos x\right)\sqrt{\cos4x}}dx$
$\bg_white I=\int_{0}^{\frac{\pi}{8}}\frac{\left(1-\sin{2x}\right)\ln\left(\tan2x\right)}{-\cos{2x}\sqrt{\cos4x}}dx$
Let $\bg_white t=\tan{2x}$
$\bg_white I=-\frac{1}{2}\int_{0}^{1}\frac{\left(1-\frac{t}{\sqrt{1+t^2}}\right)\ln{t}}{\frac{1}{\sqrt{1+t^2}}\sqrt{\frac{1-t^2}{1+t^2}}}\cdot\frac{1}{1+t^2}dt$
$\bg_white I=-\frac{1}{2}\int_{0}^{1}\frac{1}{\sqrt{1-t^2}}\ln{t}-\frac{t}{\sqrt{1-t^4}}\ln{t}dt$
Let $\bg_white J=\int_{0}^{1}\frac{\ln{t}}{\sqrt{1-t^2}}dt$ and $\bg_white K=\int_{0}^{1}\frac{t\ln{t}}{\sqrt{1-t^4}}dt$

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

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

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

#### stupid_girl

##### Active Member
$\bg_white \int_{0}^{\pi}\left(3\pi-\cos x-2x\right)\left(\frac{x}{1+\sin x}\right)^{2}dx$

#### sharky564

##### Member
$\bg_white \int_0^{\frac{1}{2}} \frac{\tan^{-1}\left ( x^3 + \frac{3x}{4}\right ) + 2 \tan^{-1}\left (e^{\frac{1}{3} \sinh^{-1}(4x)} \right )}{1 + x^2} \ \mathrm{d}x = \cot^{-1}(2)^2 + \frac{\pi}{2}\cot^{-1}(2)$

#### Canteen

##### New Member
$\bg_white \int_{0}^{\pi}\left(3\pi-\cos x-2x\right)\left(\frac{x}{1+\sin x}\right)^{2}dx$
To get this thread going again I thought I'd give one of these a go.

Letting $\bg_white u = \tfrac{\pi}{2} - x$:
\bg_white \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 $\bg_white [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, so they go to 0. We end up with:

\bg_white \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 $\bg_white \tfrac{2\pi^3}{3} + \pi^2 - 2\pi$.

#### chilli 412

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

Letting $\bg_white u = \tfrac{\pi}{2} - x$:
\bg_white \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 $\bg_white [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, so they go to 0. We end up with:

\bg_white \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 $\bg_white \tfrac{2\pi^3}{3} + \pi^2 - 2\pi$.
what was your thought process behind that u-sub?