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Hard integrals
Entrepreneur   27
N Nov 16, 2024 by soryn
Source: Own
$$\color{blue}{\int \sqrt{\cot x}dx}$$$$\color{blue}{\int \tan^2x\sec xdx}$$$$\color{blue}{\int \sec^3xdx}$$
27 replies
Entrepreneur
Aug 12, 2024
soryn
Nov 16, 2024
Hard integrals
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Source: Own
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Entrepreneur
1078 posts
#1 • 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 2 times. Last edited by Entrepreneur, Aug 31, 2024, 7:48 PM
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alexheinis
10261 posts
#3
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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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Entrepreneur
1078 posts
#4
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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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Entrepreneur
1078 posts
#5
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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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Entrepreneur
1078 posts
#6
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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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Entrepreneur
1078 posts
#7
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$$\color{blue}{\int_0^{\infty}\sqrt[3]{\frac{\sin^2x}{1+x^4}}dx }$$
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Entrepreneur
1078 posts
#8
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$$\textcolor{blue}{\int \sqrt{\tanh x}dx}$$
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naenaendr
561 posts
#9
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Entrepreneur wrote:
$$\textcolor{blue}{\int \sqrt{\tanh x}dx}$$

nice one!
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Entrepreneur
1078 posts
#10
Y by
$$\textcolor{blue}{\int_0^{\frac{\pi}{2}}\frac{\ln \cos x}{\tan x}\ln\left(\frac{\ln \cos x}{\ln \sin x}\right)dx}$$
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Entrepreneur
1078 posts
#11
Y by
$$\color{blue}{\int_0^{\frac{\pi}{2}}\ln\left(\frac{1+\tan x}{1-\tan x}\right)dx}$$$$\color{blue}{\int_0^{\frac{\pi}{2}}x\ln\left(\frac{1+\tan x}{1-\tan x}\right)dx}$$$$\color{blue}{\int_0^{\frac{\pi}{2}}x\cos x\ln\left(\frac{1+\tan x}{1-\tan x}\right)dx}$$
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Aiden-1089
231 posts
#12
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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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Entrepreneur
1078 posts
#14
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$$\color{blue}{\int_0^1\frac{\ln x}{1+x^2}dx=-G}$$
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Entrepreneur
1078 posts
#15
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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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Entrepreneur
1078 posts
#16
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$$\color{blue}{\int_0^{\infty}\frac{x^3\ln x}{(1+x^2)^2}dx}$$
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Entrepreneur
1078 posts
#17
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$$\color{blue}{\int \frac{x\ln x}{(1+x^2)^2}dx}$$
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Entrepreneur
1078 posts
#18
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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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Entrepreneur
1078 posts
#19
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$$\textcolor{blue}{\int \sqrt[5]{1+e^{\sin x}}dx}$$
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CAS03
135 posts
#20
Y by
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
135 posts
#21
Y by
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 :)
This post has been edited 1 time. Last edited by CAS03, Sep 9, 2024, 6:56 PM
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Entrepreneur
1078 posts
#22
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Click to reveal hidden text My pleasure :)
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Entrepreneur
1078 posts
#23
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$$\color{blue}{\int \frac{\tan x}{\sqrt{1+x^2}}dx}$$$$\color{blue}{\int \frac{\arctan x}{\sqrt{1+x^2}}dx}$$$$\color{blue}{\int \frac{\tanh x}{\sqrt{1+x^2}}dx}$$
This post has been edited 1 time. Last edited by Entrepreneur, Sep 13, 2024, 6:51 PM
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Entrepreneur
1078 posts
#24
Y by
$$\textcolor{blue}{\int \frac{\cot^{-1}\sqrt{1+2x}}{x\sqrt{1+2x}}dx}$$$$\textcolor{blue}{\int \frac{\cot^{-1}x}{x^2-1}dx}$$
This post has been edited 1 time. Last edited by Entrepreneur, Sep 14, 2024, 12:14 PM
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Entrepreneur
1078 posts
#25
Y by
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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Entrepreneur
1078 posts
#26
Y by
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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Entrepreneur
1078 posts
#27
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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Entrepreneur
1078 posts
#28
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$$\textcolor{blue}{\int x^n\ln^axdx=\frac{x^n\ln^ax}{n+1}-\frac{x^n}{n+1}\sum_{k=1}^{a}\frac{\ln^{a-k}x}{(n+1)^k}+\cal C.}$$Is this result true?
This post has been edited 1 time. Last edited by Entrepreneur, Nov 16, 2024, 10:01 AM
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Entrepreneur
1078 posts
#29
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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
5189 posts
#30
Y by
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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