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Source: Own, some inspired

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#1 h Aug 12, 2024, 4:23 PM • 2 Y

Y by manlio, w32dedorh30

$\color{blue}{\int \sqrt{\cot x}dx}$ $\color{blue}{\int \tan^2x\sec xdx}$ $\color{blue}{\int \sec^3xdx}$

This post has been edited 3 times. Last edited by Entrepreneur, Dec 1, 2024, 7:12 PM

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#3 h Aug 13, 2024, 9:18 AM

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$\int \sqrt{\cot x}dx=-\int {{\sqrt{p}dp}\over {p^2+1}}=-2\int{{q^2 dq}\over {q^4+1}}$ . Now factor the numerator as $(q^2-q\sqrt{2}+1)(q^2+q\sqrt{2}+1)$ and we get ${{-1}\over {2\sqrt{2}}}(\ln ({{q^2+1-q\sqrt{2}}\over {q^2+1+q\sqrt{2}}})+2\arctan (1+q\sqrt{2})-2\arctan (1-q\sqrt{2}))$ .

$\int {{\sin^2 x dx}\over {\cos^3x}}=\int {{\sin^2 x d(\sin x)}\over {(1-\sin^2 x)^2}}=\int {{p^2 dp}\over {(1-p^2)^2}}={1\over 4}(\ln ({{1+p}\over {1-p}})+{{2p}\over {1-p^2}}$ ). After this we can write ${{2p}\over {1-p^2}}={{2\sin x}\over {\cos^2 x}}$ and we can use ${{1+\cos t}\over {1-\cos t}}=\cot^2 (t/2)$ . A bit tedious so I'll skip that.

$\int{{dx}\over {\cos^3 x}}=\int {{d(\sin x)}\over {(1-\sin^2 x)^2}}=\int {{dp}\over {(1-p^2)^2}}={1\over 4}(\ln ({{1+p}\over {1-p}})+{{2p}\over {1-p^2}})$ .

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#4 h Aug 13, 2024, 3:22 PM

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$\color{blue}{\int \frac{dx}{\sqrt{\sin(1+\cos x)}}}$ $\color{blue}{\int \frac{dx}{\sqrt{\sin(1+\cos^3x)}}}$

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#5 h Aug 13, 2024, 3:28 PM

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Entrepreneur wrote:

$\color{blue}{\int \frac{dx}{\sqrt{\sin(1+\cos x)}}}$

Substitute $t=\tan \frac x2$ to obtain
$\begin{align*} \mathcal I&= \int \frac{dx}{\sqrt{\sin x(1+\cos x)}}\\ &=\int \frac{dt}{\sqrt{t}}\\ &=2\sqrt t+\mathcal C\\ &=\boxed{2\sqrt{\tan \frac x2}+\mathcal C.} \end{align*}$

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1170 posts

#6 h Aug 17, 2024, 6:38 AM

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$\textcolor{blue}{\int \frac{\sqrt{e^x}\cos x}{\sqrt[3]{3\cos x+4\sin x}}dx}$

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1170 posts

#7 h Aug 18, 2024, 10:41 AM

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$\color{blue}{\int_0^{\infty}\sqrt[3]{\frac{\sin^2x}{1+x^4}}dx }$

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1170 posts

#8 h Aug 27, 2024, 6:37 AM

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$\textcolor{blue}{\int \sqrt{\tanh x}dx}$

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#9 h Aug 27, 2024, 11:11 AM

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Entrepreneur wrote:

$\textcolor{blue}{\int \sqrt{\tanh x}dx}$

nice one!

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1170 posts

#10 h Aug 29, 2024, 3:34 PM

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$\textcolor{blue}{\int_0^{\frac{\pi}{2}}\frac{\ln \cos x}{\tan x}\ln\left(\frac{\ln \cos x}{\ln \sin x}\right)dx}$

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1170 posts

#11 h Aug 30, 2024, 1:01 PM

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$\color{blue}{\int_0^{\frac{\pi}{2}}\ln\left(\frac{1+\tan x}{1-\tan x}\right)dx}$ $\color{blue}{\int_0^{\frac{\pi}{2}}x\ln\left(\frac{1+\tan x}{1-\tan x}\right)dx}$ $\color{blue}{\int_0^{\frac{\pi}{2}}x\cos x\ln\left(\frac{1+\tan x}{1-\tan x}\right)dx}$

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#12 h Aug 30, 2024, 5:33 PM

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For $x \in (\frac{\pi}{4}, \frac{\pi}{2})$ , we have $\frac{1+\tan{x}}{1-\tan{x}} < 0$ , so $\ln \left( \frac{1+\tan{x}}{1-\tan{x}} \right)$ is undefined.

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1170 posts

#14 h Aug 31, 2024, 7:49 PM

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$\color{blue}{\int_0^1\frac{\ln x}{1+x^2}dx=-G}$

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1170 posts

#15 h Sep 5, 2024, 1:27 PM

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$\color{blue}{\int \sqrt{e^{2x}+e^x+1}dx}$ $\color{blue}{\int \sqrt{e^{3x}+e^x+1}dx}$ $\color{blue}{\int \frac{1}{\sqrt{e^{3x}+e^x+1}}dx}$

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#16 h Sep 5, 2024, 6:37 PM

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$\color{blue}{\int_0^{\infty}\frac{x^3\ln x}{(1+x^2)^2}dx}$

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1170 posts

#17 h Sep 5, 2024, 6:54 PM

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$\color{blue}{\int \frac{x\ln x}{(1+x^2)^2}dx}$

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#18 h Sep 5, 2024, 7:06 PM

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Entrepreneur wrote:

$\color{blue}{\int \frac{x\ln x}{(1+x^2)^2}dx}$

We use $\textcolor{red}{\textit{integration by parts.}}$ Let $u=\ln x, dv=\frac{xdx}{(1+x^2)^2}.$ Then, $du=\frac 1x,v=\frac{-1}{2(1+x^2)}.$ Thus,
$\mathcal I =uv-\int vdu=\frac{-\ln x}{2(1+x^2)}+\frac 12\color{magenta}{\int\frac{dx}{x(1+x^2)}}.$ Now, for the magenta integral, let $x=\tan \theta.$ Then, $dx=\sec^2\theta d\theta.$ So, $\color{magenta}{\int\frac{dx}{x(1+x^2)}}=\int\cot \theta d\theta =\ln|\sin \theta|+\mathcal C=\ln\left | \frac{x}{\sqrt{1+x^2}}\right |+\mathcal C.$ Hence, $\boxed{\int\frac{x\ln x}{(1+x^2)^2}dx=\frac 12\ln\left | \frac{x}{\sqrt{1+x^2}}\right |-\frac{\ln x}{2(1+x^2)}+\mathcal C.}$

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1170 posts

#19 h Sep 9, 2024, 12:08 PM

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$\textcolor{blue}{\int \sqrt[5]{1+e^{\sin x}}dx}$

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213 posts

CAS03

#20 h Sep 9, 2024, 6:55 PM

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Entrepreneur wrote:

$\color{blue}{\int \sec^3xdx}$

Haven't seen much on this one and it actually comes together very nicely

Separate:
$\int \sec^3 x \, dx = \int \sec x \cdot \sec^2 x \, dx.$ Integration by parts:
$\begin{array}{cc} u=\sec x & dv = \sec^2 x \, dx \\ du = \sec x \tan x \, dx & v = \tan x \end{array}$ Substituting:
$\begin{align*} \int \sec^3x \, dx &= \sec x \tan x - \int \sec x \tan^2 \, dx \\ \int \sec^3x \, dx &= \sec x \tan x - \int \sec x (\sec^2 x -1) \, dx \\ 2\int \sec^3 x \, dx &= \sec x \tan x + \int \sec x \, dx \end{align*}$ So final answer:
$\boxed{\int \sec^3x \, dx = \frac 12 \left( \sec x \tan x + \log |\sec x + \tan x | + C \right)}.$

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CAS03

#21 h Sep 9, 2024, 6:56 PM

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Thanks for sending these, they are pretty cool and fun especially as I literally just learned integration by parts this week in class lol

That one at least was good practice

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#22 h Sep 10, 2024, 7:32 AM

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Click to reveal hidden text My pleasure

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#23 h Sep 13, 2024, 6:50 PM

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$\color{blue}{\int \frac{\tan x}{\sqrt{1+x^2}}dx}$ $\color{blue}{\int \frac{\arctan x}{\sqrt{1+x^2}}dx}$ $\color{blue}{\int \frac{\tanh x}{\sqrt{1+x^2}}dx}$

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#24 h Sep 14, 2024, 12:04 PM

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$\textcolor{blue}{\int \frac{\cot^{-1}\sqrt{1+2x}}{x\sqrt{1+2x}}dx}$ $\textcolor{blue}{\int \frac{\cot^{-1}x}{x^2-1}dx}$

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#25 h Sep 22, 2024, 9:39 PM

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Entrepreneur wrote:

$\color{blue}{\int_0^1\frac{\ln x}{1+x^2}dx=-G}$

$G$ is the Catalan's constant.

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#26 h Oct 13, 2024, 2:53 PM

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Entrepreneur wrote:

$\color{blue}{\int_0^1\frac{\ln x}{1+x^2}dx=-G}$

We have
$\begin{align*} \mathcal I&= \int_0^1\frac{\ln x}{1+x^2}dx\\ &=\int_0^1\ln x\sum_{k=0}^\infty(-1)^kx^{2k}dx\\ &=\sum_{k=0}^\infty\int_0^1x^{2k}\ln xdx\\ &=\boxed{\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+1)^2} =-G.} \end{align*}$ $\color{green}{\cal{QED.}}$

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1170 posts

#27 h Oct 14, 2024, 12:11 PM

Y by

Entrepreneur wrote:

Entrepreneur wrote:

$\color{blue}{\int_0^1\frac{\ln x}{1+x^2}dx=-G}$

We have
$\begin{align*} \mathcal I&= \int_0^1\frac{\ln x}{1+x^2}dx\\ &=\int_0^1\ln x\sum_{k=0}^\infty(-1)^kx^{2k}dx\\ &=\sum_{k=0}^\infty\int_0^1x^{2k}\ln xdx\\ &=\boxed{\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+1)^2} =-G.} \end{align*}$ $\color{green}{\cal{QED.}}$

Generalized Result:
$\color{blue}{\int_0^1\frac{\ln x}{x^a\pm1}dx=\sum_{k=0}^\infty\frac{(\mp 1)^{k+1}}{(ak+1)^2}.}$
Can this be generalized any further?

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1170 posts

#28 h Nov 13, 2024, 10:42 AM

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$\color{blue}{\int_0^1x^n\ln^m xdx=\frac{(-1)^m \Gamma(m+1)}{(n+1)^{m+1}}.}$ $\color{blue}{\int_0^1\frac{x^{p-1}\ln^m x}{(1+x^q)^n}dx=(-1)^m\Gamma(m+1)\sum_{k=0}^\infty\frac{(-1)^{k}}{(p+kq)^{m+1}}{{n+k-1}\choose {k}}.}$ $\color{blue}{\int_0^1\ln^n\left(\frac{1-x}{1+x}\right)dx=(-1)^n\cdot 2\cdot n!\cdot \eta(n).}$ $\color{blue}{\int_0^1x\ln^n\left(\frac{1-x}{1+x}\right)dx=(-1)^n\cdot 2\cdot n!\cdot \eta(n-1).}$ $\color{blue}{\int_0^1x^2\ln^n\left(\frac{1-x}{1+x}\right)dx=\frac{(-1)^n\cdot 2\cdot n!}{3}\Big(2\eta(n-2)+\eta(n)\Big).}$ $\color{blue}{\int_0^1\frac{\ln^n x}{(1+x)^5}dx=\frac{(-1)^n\cdot n!}{24}\Big(\eta(n-3)+6\eta(n-2)+11\eta(n-1)+6\eta(n)\Big).}$ $\color{blue}{\int_0^1\frac{\ln^n x}{(1+x)^6}dx=\frac{(-1)^n\cdot n!}{120}\Big(\eta(n-4)+10\eta(n-3)+33\eta(n-2)+50\eta(n-1)+24\eta(n)\Big).}$

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#29 h Nov 16, 2024, 10:01 AM

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$\color{blue}{\int_0^\frac{\pi}{4}\ln(\cos x)dx}$ $\color{blue}{\int_0^\frac{\pi}{4}\ln^2(\cos x)dx}$

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soryn

#30 h Nov 16, 2024, 12:55 PM

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For the first integral,denote I the desired integral,and J the integral with ln(sinx). After calculatios,obtain the system with the equations : I+J=-π/2ln2 and I-J=G, where G is the Catalan's constant.

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#31 h Nov 29, 2024, 5:58 PM

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$\color{blue}{\int_{-\infty}^{\infty}\frac{\sec x}{x^2+a^2}dx.}$ $\color{blue}{\int_{-\infty}^{\infty}\frac{\sec(bx)}{x^2+a^2}dx.}$

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#32 h Nov 30, 2024, 11:38 AM

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$\color{blue}{\int_0^1\sqrt{\ln\left(\frac{1-x}{1-x}\right)}dx.}$ $\color{blue}{\int_0^1\sqrt{\ln^3\left(\frac{1-x}{1+x}\right)}dx.}$

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#33 h Dec 1, 2024, 8:35 AM

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Entrepreneur wrote:

$\color{blue}{\int_0^1x^n\ln^m xdx=\frac{(-1)^m \Gamma(m+1)}{(n+1)^{m+1}}.}$ $\color{blue}{\int_0^1\frac{x^{p-1}\ln^m x}{(1+x^q)^n}dx=(-1)^m\Gamma(m+1)\sum_{k=0}^\infty\frac{(-1)^{k}}{(p+kq)^{m+1}}{{n+k-1}\choose {k}}.}$ $\color{blue}{\int_0^1\ln^n\left(\frac{1-x}{1+x}\right)dx=(-1)^n\cdot 2\cdot n!\cdot \eta(n).}$ $\color{blue}{\int_0^1x\ln^n\left(\frac{1-x}{1+x}\right)dx=(-1)^n\cdot 2\cdot n!\cdot \eta(n-1).}$ $\color{blue}{\int_0^1x^2\ln^n\left(\frac{1-x}{1+x}\right)dx=\frac{(-1)^n\cdot 2\cdot n!}{3}\Big(2\eta(n-2)+\eta(n)\Big).}$ $\color{blue}{\int_0^1\frac{\ln^n x}{(1+x)^5}dx=\frac{(-1)^n\cdot n!}{24}\Big(\eta(n-3)+6\eta(n-2)+11\eta(n-1)+6\eta(n)\Big).}$ $\color{blue}{\int_0^1\frac{\ln^n x}{(1+x)^6}dx=\frac{(-1)^n\cdot n!}{120}\Big(\eta(n-4)+10\eta(n-3)+33\eta(n-2)+50\eta(n-1)+24\eta(n)\Big).}$

For first, we substitute $x=e^{\frac{-u}{n+1}},$ so that $dx=\frac{-e^{\frac{-u}{n+1}}du}{n+1}.$ Therefore,
$\begin{align*} \int_0^1x^n\ln^mxdx&=\int_\infty^0e^\frac{-un}{n+1}\frac{(-u)^m}{(n+1)^{m+1}}\frac{-e^\frac{-u}{n+1}}{n+1}du\\ &=\frac{(-1)^m}{(n+1)^{m+1}}\int_0^\infty u^m e^{-u}du\\ &=\boxed{\frac{(-1)^m \Gamma(m+1)}{(n+1)^{m+1}}.} \end{align*}$
For second, we use series expansion and the previous result.
$\begin{align*} \int _0^1\frac{x^{p-1}\ln^mx}{(1+x^q)^n}dx&=\int_0^1 x^{p-1}\ln^mx\sum_{k=0}^\infty (-1)^k x^{kq}{{n+k-1}\choose{k}}dx\\ &=\sum_{k=0}^\infty(-1)^k{{n+k-1}\choose{k}}\int_0^1 x^{p+kq-1}\ln^mxdx\\ &=\sum_{k=0}^\infty{{n+k-1}\choose{k}}\frac{(-1)^m\Gamma(m+1)}{(p+kq)^{m+1}}\\ &=\boxed{(-1)^m\Gamma(m+1)\sum_{k=0}^\infty\frac{(-1)^k}{(p+kq)^{m+1}}{{n+k-1}\choose{k}}.} \end{align*}$
For third, fourth and fifth, use the previous results combined with the substitution $t=\frac{1-x}{1+x},$ so that $x=\frac{1-t}{1+t},\;dx=\frac{-2dt}{(1+t)^2}.$ Therefore,
$\begin{align*} \int_0^1\ln^n\left(\frac{1-x}{1+x}\right)dx&=\int_1^0\ln^n t\frac{-2dt}{(1+t)^2}\\ &=2\int_0^1\frac{\ln^ntdt}{(1+t)^2}\\ &=2\cdot (-1)^n\Gamma(n+1)\sum_{k=0}^\infty\frac{(-1)^k}{(1+k)^n}\\ &=\boxed{2\cdot(-1)^n\cdot n!\cdot \eta(n).} \end{align*}$ Similarly,
$\begin{align*} \int_0^1x\ln^n\left(\frac{1-x}{1+x}\right)dx&=2\int_0^1\frac{(1-t)\ln^nt}{(1+t)^3}dt\\ &=4\int_0^1\frac{\ln^ntdt}{(1+t)^3}-2\int_0^1\frac{\ln^ntdt}{(1+t)^2}\\ &=\boxed{2\cdot(-1)^n\cdot n!\cdot \eta(n-1).} \end{align*}$ And, finally
$\begin{align*} \int_0^1x^2\ln^n\left(\frac{1-x}{1+x}\right)dx&=2\int_0^1\frac{(1-t)^2\ln^ntdt}{(1+t)^4}\\ &=8\int_0^1\frac{\ln^ntdt}{(1+t)^4}-8\int_0^1\frac{\ln^ntdt}{(1+t)^3}+2\int_0^1\frac{\ln^ntdt}{(1+t)^2}\\ &=\boxed{\frac{(-1)^n\cdot 2\cdot n!}{3}\Big(2\eta(n-2)+\eta(n)\Big).} \end{align*}$

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#35 h Jan 5, 2025, 3:02 PM

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$\color{blue}{\int\frac{dx}{(1+a^2x^2)(1+b^2x^2)(1+c^2x^2)}.}$

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