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These are some of the integrals with beautiful solutions I came across-

$$\int \frac{x^2}{(x\sin x+\cos x)^2} dx$$

$$\int\frac {1}{\sin^3x+\cos^3x} dx$$

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

I'd love if you share some of the ones you came across.

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closed as too broad by Antonio Vargas, Hans Engler, Dan Rust, Grigory M, Emily May 15 '14 at 19:42

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  • $\begingroup$ Define "beautiful". $\endgroup$ – evil999man May 15 '14 at 17:50
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    $\begingroup$ I know perception of beauty is subjective, but what do you see as beautiful about the solution of the third integral, for example? $\endgroup$ – mirgee May 15 '14 at 17:51
  • $\begingroup$ @mirgee actually, second one integrates pretty ugly too. $\endgroup$ – Kaster May 15 '14 at 17:53
  • $\begingroup$ I've always been very fond of the integrals of $\sec{x}$ and $\sec^3{x}$. $\endgroup$ – David H May 15 '14 at 17:53
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    $\begingroup$ I believe this should be a community wiki though... $\endgroup$ – Dair May 15 '14 at 17:56
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\begin{align} I_1 & = \int \sqrt{ \sqrt{ x + 2\sqrt{2x-4} } + \sqrt{ x - 2\sqrt{2x-4} } } \,\mathrm{d}x \, , \quad x>4\\ I_2 & = \int \log( \log x) + \frac{2}{\log x} - \frac{1}{(\log x)^2} \mathrm{d}x \\ I_4 & = \int (1 + 2x^2) e^{x^2}\, \mathrm{d}x \\ I_5 & = \int \frac{\sqrt{x+\sqrt{x^2+1\,}\,}\,}{\sqrt{x^2+1\,}\,} \mathrm{d}x \\ I_6 & = \int \frac{2^x 3^x}{9^x - 4^x} \,\mathrm{d}x \end{align} \begin{align*} I_7 = \int \left( \frac{\arctan x}{x - \arctan x}\right)^2 \mathrm{d}x = \frac{1 + x \arctan x}{\arctan x - x} = \frac{1}{\tan (\beta - \tan \beta)}\,, \end{align*} where $x = \tan \tan \beta$ or $\beta = \arctan (\arctan x)$. $$ I_6 = \int \frac{x^2+2x+1+ (3x+1)\sqrt{x+\ln x}}{x\,\sqrt{x+\ln x}(x+\sqrt{x+\ln x})}\mathrm{d}x = 2 (\sqrt{x+\ln x} + \ln(x+\sqrt{x+\ln x})) + C $$ I have a bunch more of these here, see p.68 for instance. (click on the problems for solution)

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  • $\begingroup$ awesome work, thank you for posting this! $\endgroup$ – userX May 16 '14 at 16:32
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This isn't indefinite. But it's crazy

$$ \int_0^{\pi/2} \frac{ d \theta}{\sqrt{a^2\cos^2\theta +b^2 \sin^2\theta }} = \frac{\pi}{2AGM(a,b)} $$

Where AGM is the arithmetic geometric mean.

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$$\int\dfrac{x^{4n}(1+x^{4n})}{1+x^2}dx$$

Why? Because from $0$ to $1$ they give good approximations of $\pi$. See this

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    $\begingroup$ cleverly constructed!! (+1) $\endgroup$ – Santosh Linkha May 15 '14 at 18:09
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$$\int (\sqrt {\tan x}+\sqrt{\cot x})dx=\sqrt 2\arctan\dfrac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt 2} +C$$

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Wolfram alpha gives time exceeded on this one :

$$\int\dfrac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx=\sqrt{t^2+2t-3}-\ln{(t+1+\sqrt{t^2+2t-3})}+\sqrt 3 \arcsin{\dfrac{t+5}{2(t+2)}}+C$$

where $t=x+\dfrac 1 x$

One does not simply integrate this.

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  • $\begingroup$ I don't understand the reason of the downvote. '+1' $\endgroup$ – user103816 Jun 3 '14 at 16:20
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$$\int \left| \sin{ax} \right|\,dx = {2 \over a} \left\lfloor \frac{ax}{\pi} \right\rfloor - {1 \over a} \cos{\left( ax - \left\lfloor \frac{ax}{\pi} \right\rfloor \pi \right)} + C$$

$$\int \left|\cos {ax}\right|\,dx = {2 \over a} \left\lfloor \frac{ax}{\pi} + \frac12 \right\rfloor + {1 \over a} \sin{\left( ax - \left\lfloor \frac{ax}{\pi} + \frac12 \right\rfloor \pi \right)} + C$$

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The Gaussian integral isn't indefinite but its derivation and answer are still remarkable:

$$\int_{-\infty}^{+\infty} e^{-x^2}\,dx = \sqrt{\pi}$$

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