Past Problems List (Integration Contest)

Season 1:

1:0alnxaxx2dxsin(2015x)sin2013xdx0(sin(x)x)2dx0π/2ln(cosx)sinxdx0dx1+xn\displaystyle 1:\int_0^a \frac{\ln x}{\sqrt{ax-x^2}}dx\quad\int\sin(2015 x)\sin^{2013}xdx\quad\int _0^\infty{(\frac { sin(x) }{ x } })^2dx\quad\int_0^{\pi/2}\frac{\ln(\cos x)}{\sin{x}}dx\quad\int_0^ \infty\frac{dx}{1+x^n}

6:0π/2sin2xsinx+cosxdx0π1(10+cosx)30π2dx1+8sin2(tanx)01ln(1+x1x)dxx1x20sin3xx2dx\displaystyle 6:\int_0^{\pi/2}\frac{\sin^2x}{\sin x+\cos x}dx\quad\int_0^\pi\dfrac{1}{(\sqrt{10}+ \cos x)^3}\quad\int_0^{\frac{\pi}{2}}\frac{dx}{1+8\sin^2(\tan x)}\quad\int_0^1\ln\left(\frac{1+x}{1-x}\right)\frac{dx}{x\sqrt{1-x^2}}\int_0^\infty\frac{\sin^3x}{x^2}dx

11:0logxx2+a2dxa>0cos(arctan2x)(1+x2)1+4x2dx0xnex1dx01ln(3+x3x)dxx(1x)011lnx+11xdx\displaystyle 11:\int_{0}^{\infty}\dfrac{\log{x}}{x^2+a^2}dx\,a>0\quad\int_{-\infty}^\infty\frac{\cos (\arctan 2x)}{(1+x^2)\sqrt{1+4x^2}}dx\quad\int_0^\infty\frac{x^n}{e^x-1}dx\quad\int_0^1 \ln(\frac{3+x}{3-x})\frac{dx}{\sqrt{x(1-x)}}\quad\int_0^1\frac{1}{\ln{x}}+\frac{1}{1-x}dx

16:0π2cosn1xcosaxdx11dx(1x)(1+x)20π/4tan1/3xdx1313x41x4cos1(2x1+x2)dx\displaystyle 16:\int_0^{\frac\pi2}\cos^{n-1}x\cos{ax}dx\quad\int_{-1}^1\frac{dx}{\sqrt{(1-x)(1+x)^2}}\quad\int_0^{\pi/4}\tan^{1/3}xdx\quad\int_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\frac{x^4}{1-x^4}\cos^{-1}{(\frac{-2x}{1+x^2})}dx

20:01sinh1(x)log((21)x+1)xdx0dxx4+2cos(2θ)x2+10axdxcos(x)cos(ax)0π4lntan(x)dx\displaystyle 20:\int_0^1\frac{\sinh^{-1}(x)-\log((\sqrt{2}-1)\sqrt{x}+1)}{x}dx\quad\int_0^\infty\frac{dx}{x^4+2\cos(2\theta)x^2+1}\quad\int_0^a \frac{xdx}{\cos(x)\cos(a-x)}\quad\int_0^{\frac{\pi}{4}}\ln{\tan(x)}dx

24:01arcsechxarcsinxdx01ln(ln(1x))dx0lnxcoshxdx0ln2tanhxdxsinh2xcos2xsinhπxdx\displaystyle 24:\int_0^1\operatorname{arcsech}x\arcsin{x}dx\quad\int_0^1ln(ln(\frac{1}{x}))dx\quad\int_0^\infty\frac{\ln{x}}{\cosh{x}}dx\quad\int_0^\infty\ln^2{\tanh{x}}dx\quad\int_{-\infty}^\infty\frac{\sinh{2x}\cos{2x}}{\sinh{\pi x}}dx

29:01x2arctanx2dx1xx0.5xdx0πln(12acosx+a2)dxxsin(x+sinx)ecosx1+x2dx\displaystyle 29:\int_0^1x^2\arctan{x^2}dx\quad\int_1^\infty\frac{x-\left\lfloor x\right\rfloor-0.5}{x}dx\quad\int_0^\pi\ln{(1-2a\cos{x}+a^2)}dx\quad\int_{-\infty}^\infty\frac{x\sin(x+\sin x)e^{\cos x}}{1+x^2}dx

33:0tlog(2t+1)(t+1)(2t+1)dtcos(sarctan(ax))(1+x2)(1+a2x2)s/2dx0arctan(x2x2+1)x4+1dx36A:11xa1+ebxdx\displaystyle 33:\int_0^\infty\frac{\sqrt{t}\log(2t+1)}{(t+1)(2t+1)}dt\quad\int_{-\infty}^\infty\frac{\cos(s\arctan(ax))}{(1+x^2)(1+a^2x^2)^{s/2}}dx\quad\int_0^\infty\frac{\arctan(\frac{x^2}{x^2+1})}{x^4+1}dx\quad 36A:\int_{-1}^1\frac{x^a}{1+e^{bx}}dx

36B:0π2tanxsecx(tanx+secx)ndx01lnxln(1x)dx0π2+2cos(x)cos((281)x)2cos(28x)cos((28+1)x)1cos(2x)\displaystyle 36B:\int_0^\frac{\pi}{2}\frac{\tan{x}\sec{x}}{(\tan x+\sec x)^n}dx\quad\int_0^1\ln{x}ln(1-x)dx\quad\int_0^\pi\frac{2+2\cos(x)-\cos((2^{8}-1)x)-2\cos(2^{8}x)-\cos((2^8+1)x)}{1-\cos(2x)}

39:0arctan(x)arctan(2x)x2dxcos(ax2)sin(ax2)1+x4dx0π/12ln(tan(x))dx011xlnxxdx\displaystyle 39:\int_0^\infty\frac{\arctan(x)\arctan(2x)}{x^2}dx\quad\int_{-\infty}^\infty\frac{\cos(ax^2)-\sin(ax^2)}{1+x^4}dx\quad\int_{0}^{\pi/12} \ln(\tan(x)) dx\quad\int_0^1\frac{1}{x}\frac{\ln{x}}{\sqrt{x}}dx

43:0π21+sin2xdx0sinh(ax)sin(bx)(cosh(ax)+cos(bx))2dx\displaystyle 43:\int_0^\frac{\pi}{2}\sqrt{1+\sin^2{x}}dx\quad\int_0^\infty\frac{\sinh(ax)\sin(bx)}{(\cosh(ax)+\cos(bx))^2}dx

Season 2:

1:01xlnx1x2dx0xcos(x3)ex3dx0π2tan35xdx01arctan(P(x))arctanx1xdx0π/2arctan1729cos2xdx\displaystyle 1:\int_0^1\frac{x\ln{x}}{\sqrt{1-x^2}}dx \quad\int_0^\infty\frac{x\cos(x^3)}{e^{x^3}}dx\quad\int_0^\frac{\pi}{2}\tan^\frac{3}{5} {x}dx\quad\int_0^1\arctan(P(x))\arctan{\sqrt{\frac{x}{1-x}}}dx\quad\int_0^{\pi /2}\arctan{\frac{1729}{cos^2{x}}}dx

6:abarccos(x(a+b)xab)dx0[xv3(γx+logΓ(1+x))]dx12πea2t2eiωtdtx2+20(xsinx+5cosx)2dx\displaystyle 6:\int_a^b\arccos(\frac{x}{\sqrt{(a+b)x - ab}})dx\quad\int_0^\infty [x^{v-3}(\gamma x+\log\Gamma (1+x))]dx\quad\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-a^2t^2}e^{i\omega t}dt\int\frac{x^2+20}{(x\sin{x}+5\cos{x})^2}dx

10:0π21(9tan2x+16)3dx01ln(x+1)(x+1)(x2+1)dx0π2sina1(θ)cos2ta(θ)sin(2tθ)dθ01ln(1+x1x)dx\displaystyle 10:\int_0^\frac{\pi}{2}\frac{1}{(9\tan^2{x}+16)^3}dx\quad\int_0^1\frac{\ln{(x+1)}}{(x+1)(x^2+1)}dx\int_0^\frac{\pi}{2}\sin^{a-1}{(\theta)}\cos^{2t-a}{(\theta)}\sin{(2t\theta)}d\theta\quad\int_0^1\ln(\frac{1+x}{1-x})dx

14:011xlnx(x+x2+x4+x8+)dxπ2π2cos(xtan(θ))dθ01tarccost1+t4dtx2x4x3+x2x+1dx\displaystyle 14:\int^1_0\frac{1-x}{\ln{x}}(x+x^2+x^{4}+x^{8}+\cdots )dx\quad\int_\frac{-\pi}{2}^\frac{\pi}{2}\cos(x\tan(\theta))d\theta\quad\int^1_0\frac{t\arccos{t}}{1+t^4}dt\quad\int_{-\infty}^\infty\frac{x^2}{x^4-x^3+x^2-x+1}dx

18:0log2t1+t2dt0π3ln2(sinxsin(π3+x))dx0dx(1+x)3+1π2π2ln(1+bsinx)sinxdx0πxsinx(cos2x+3)2dx\displaystyle 18:\int_0^\infty\frac{\log^2{t}}{1+t^2}dt\quad\int^\frac{\pi}{3}_0\ln^{2}(\frac{\sin{x}}{\sin(\frac{\pi}{3}+x)})dx\quad\int_0^\infty \frac{dx}{(1+x)^3 +1}\quad\int_\frac{-\pi}{2}^\frac{\pi}{2}\frac{\ln(1+ b\sin{x})}{\sin{x}}dx\quad\int_0^\pi\frac{x\sin{x}}{(\cos^2{x}+3)^2}dx

23:01log(1+x)log(1x3)dx0π21sin8x+cos8xdx0π2ln(sinx)ln(cosx)tanxdx0πe2cosxsin2(sinx)dx\displaystyle 23:\int_0^1\log(1+x)\log(1-x^3)dx\quad\int_0^\frac{\pi}{2}\frac1{\sin^8{x}+\cos^8{x}}dx\quad\int^\frac{\pi}{2}_0\frac{\ln(\sin{x}) \ln(\cos{x})}{\tan{x}}dx\quad\int_0^\pi e^{2\cos{x}}\sin^2{(\sin{x})}dx

27:02dx(2x2x3)1/30x5sinx(1+x2)3dx01ln(cos(πx2))x(x+1)dx0π8lntan2xdxlimn01xnx2n1xdx\displaystyle 27:\int^2_0\frac{dx}{(2x^2-x^3)^{1/3}}\quad\int_0^\infty\frac{x^5\sin{x}}{(1+x^2)^3}dx\int_0^1\frac{\ln(\cos(\frac{\pi x}{2}))}{x(x+1)}dx\quad\int^\frac{\pi}{8}_0\ln{\tan{2x}}dx\quad\lim_{n\to\infty}\int_0^1\frac{x^n-x^{2n}}{1-x}dx

32:0πt2ln2(2cos(t2))dt0arcsinexdx0(sinx2x2)(x2+x1)dx0ln(1+xa1+xb)(1+x2)lnxdx\displaystyle 32:\int_0^\pi t^2\ln^2{(2\cos(\frac{t}{2}))}dt\quad\int_0^\infty\arcsin{e^{-x}}dx\quad\int_0^\infty(\frac{\sin{x^2}}{x^2})\cdot (x^2+x-1)dx\quad\int_0^\infty\frac{\ln(\frac{1+x^a}{1+x^b})}{(1+x^2)\ln{x}}dx

36:0sinxex1dx0π2cot(θ2)cosθlncosθdθ02ln((10.25x2)3+x3)3lnx(1+0.25x2)ln(10.25x2)(1+0.25x2)lnxdx\displaystyle 36:\int_0^\infty\frac{\sin{x}}{e^x-1}dx\quad\int^\frac{\pi}{2}_0\cot(\frac{\theta}{2})\sqrt{\cos{\theta}}\ln{\cos{\theta}}d\theta\quad\int_0^2\frac{\ln((1-0.25x^2)^3+x^3)-3\ln{x}}{(1+0.25x^2)\ln(1-0.25x^2)-(1+0.25x^2)\ln{x}}dx

39:0(sinxx11+x)dxx0π2x(ln(sinx)2sinx)cotxdxx+x2+2dx0π2ln(2cos(x2))dx\displaystyle 39:\int_0^\infty(\frac{\sin{x}}{x}-\frac1{1+x})\frac{dx}{x}\quad\int_0^\frac{\pi}{2}x(\frac{\ln(\sin{x})-2}{\sqrt{\sin{x}}})\cot{x}dx\quad\int\sqrt{x+\sqrt{x^2+2}}dx\quad\int_0^\frac{\pi}{2}\ln{(2\cos(\frac{x}{2}))}dx

43:x3ex2(x2+1)2dx0ln(1+x)ln2x+π2dxx20π2sin6xtanxsintanxdxlimn0π4n(cos2nxsin2nx)tan2xdx\displaystyle 43:\int\frac{x^3e^{x^2}}{(x^2+1)^2}dx\quad\int_0^\infty\frac{\ln{(1+x)}}{\ln^2{x}+\pi^2}\frac{dx}{x^2}\quad\int_0^\frac{\pi}{2}\sin^6{x}\tan{x}\sin{\tan{x}}dx\quad\lim_{n\to\infty}\int_0^\frac{\pi}{4}n(\cos^{2n}x-\sin^{2n}x)\tan{2x}dx

47:0ex2lnxdx0π2cscxtanxsinx+cosxdx01ln(Γ(x))dx0ln(1+x)x34(1+x)dx\displaystyle 47:\int_0^\infty e^{-x^2}\ln{x}dx\quad\int_0^\frac{\pi}{2}\frac{\csc{x}\sqrt{\tan{x}}}{\sin{x}+\cos{x}}dx\quad\int_0^1\ln{(\Gamma(x))}dx\quad\int_0^\infty\frac{\ln(1+x)}{\sqrt[4]{x^3}(1+x)}dx

Season 3:

1:01cosh(alog(x))log(1+x)xdx01(x2+1)arctan3xxdx01lnxln2(1+x)xdx01Li32(x)x2dx\displaystyle 1:\int_0^1\frac{\cosh(a\log(x))\log(1+x)}{x}dx\quad\int_0^1\frac{(x^{2}+1)\arctan{3x}}{x}dx\quad\int_0^1\frac{\ln{x}\ln^2(1+x)}{x}dx\quad\int_0^1\frac{{\rm Li}_3^2(-x)}{x^2}dx

5:0ln(x2+1)arctanxeπx1dx0lnx(x+a)2+b2dx0lnxxn+1dx0π2ln2(asinθ)dθ\displaystyle 5:\int_0^\infty\frac{\ln(x^2+1)\arctan{x}}{e^{\pi x}-1}dx\quad\int_0^\infty\frac{\ln{x}}{(x+a)^2+b^2}dx\quad\int_0^\infty\frac{\ln x}{x^n+1}dx\quad\int_0^\frac{\pi}{2}\ln^2\big(a\sin\theta\big)d\theta

9:01Li22(x)dx01(x1)exlnxdx01tα1tβ1(1+t)lntdt0(lnxarctanxx)2dx02πecosθcos(nθsinθ)dθ\displaystyle 9:\int_0^1Li_2^2(x)dx\quad\int_0^1(x-1)e^{-x}\ln{x}dx\quad\int_0^1\frac{t^{\alpha-1}-t^{\beta-1}}{(1+t)\ln{t}}dt\quad \int_0^\infty(\frac{\ln{x}\arctan{x}}{x})^2dx\quad\int_0^{2\pi}e^{\cos\theta}\cos\big(n\theta-\sin\theta\big)d\theta

14:0lnx[ln(x+12)1x+1ψ(x+12)]dx\displaystyle 14:\int_0^\infty\ln{x}[\ln(\frac{x+1}{2})-\frac{1}{x+1}-\psi(\frac{x+1}{2})]dx

These are copy-pasted problems from past contests in chronological order. I will continually update with the new contests.

Note by Jasper Braun
2 years, 6 months ago

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Season 1 also has 44 more problems, 4141 to 4444.

Ishan Singh - 2 years, 6 months ago

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Thanks I've added and completed.

Jasper Braun - 2 years, 6 months ago

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Season 1 Problem 17 is typed incorrectly it should be a cube root in the denominator. As the integral is typed it diverges

D S - 2 months, 3 weeks ago

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