346. Some easy integrals.

by Virgil Nicula, Jun 20, 2012, 1:10 AM

PP1. Ascertain in the beginning three simple indefinite trigonometric integrals (with integration by parts or by substitutions):

$\left\{\begin{array}{cc} H=\int\sin 2x\ln\cos x\ \mathrm{dx} & (1)\ .\\\\ I=\int (\cos 2x + \sin 2x) \ln (\cos x+\sin x)\ \mathrm{dx} & (2)\ .\\\\ K=\int \cos 2x\ln \cos x\ \mathrm{dx} & (3)\ .\end{array}\right\|$

Proof.

$1\blacktriangleright\ H=\int\sin 2x\ln\cos x\ \mathrm{dx}=-2\cdot \int\cos x\ln\cos x(\cos x)'\ \mathrm{dx}\implies$ $\boxed{H=-2\cdot G\circ \phi}$ , where $\boxed{t=\phi (x)=\cos x}$ and $G=\int t\ln t\ \mathrm{dt}$ . Apply the integration by parts

$\boxed{\begin{array}{ccc} u(x)=\ln t & \implies & u'(x)=\frac it\\\\ v'(x)=t & \implies & v(x)=\frac {t^2}{2}\end{array}}$ and obtain that $G=\frac 12\cdot t^2\ln t-\frac 14\cdot t^2+\mathbb C$ . In conclusion, $\boxed{\int\sin 2x\ln\cos x\ \mathrm{dx}=\frac 12\cdot \cos^2x\ln\cos x-\frac 14\cdot\cos^2x+\mathbb C}$ .

$2\blacktriangleright\ I=\int (\cos 2x + \sin 2x) \ln (\cos x+\sin x)\ \mathrm{dx}$ . Apply down integration by parts $\boxed{\begin{array}{ccc} u(x)=\ln (\cos x+\sin x ) & \implies & u'(x)=\frac {\cos x-\sin x}{\cos x+\sin x}\\\\ v'(x)=\cos 2x+\sin 2x & \implies & v(x)=\frac 12\cdot\left(\sin 2x-\cos 2x\right)\end{array}}$ and obtain

$I=\frac 12\cdot (\sin 2x-\cos 2x)\ln (\cos x+\sin x)+\frac 12\cdot J$ , where $J=\int\frac {(\cos x-\sin x)(\cos 2x-\sin 2x)}{\cos x+\sin x}\ \mathrm{dx}=$

$\int\frac {(\cos x-\sin x)\left[(\cos x-\sin x)(\cos x+\sin x)-(\cos x+\sin x)^2+1\right]}{\cos x+\sin x}\ \mathrm{dx}=$ $\int\left[(\cos x-\sin x)^2-\cos 2x+\frac {\cos x-\sin x}{\cos x+\sin x}\right]\ \mathrm{dx}=$

$\int\left[1-\sin 2x-\cos 2x+\frac {\cos x-\sin x}{\cos x+\sin x}\right]\ \mathrm{dx}\implies$ $J=x+\frac 12\cdot (\cos 2x-\sin 2x)+\ln(\cos x+\sin x)+\mathbb C$ .

In conclusion, $I=\frac 12\cdot (\sin 2x+1-\cos 2x)\ln (\cos x+\sin x)+\frac 12x+\frac 14\cdot (\cos 2x-\sin 2x)+\mathbb C\implies$

$\boxed{\int (\cos 2x + \sin 2x) \ln (\cos x+\sin x)\ \mathrm{dx}=\sin x(\sin x+\cos x)\ln (\cos x+\sin x)+\frac x2+\frac 14\cdot (\cos 2x-\sin 2x)+\mathbb C}$ .

$3\blacktriangleright\ K=\int \cos 2x\ln \cos x\ \mathrm{dx}$ . Apply down integration by parts $\boxed{\begin{array}{ccc} u(x)=\ln \cos x & \implies & u'(x)=-\frac {\sin x}{\cos x}\\\\ v'(x)=\cos 2x & \implies & v(x)=\frac 12\cdot\sin 2x\end{array}}$ and obtain

$K=\frac 12\cdot \sin 2x\ln\cos x+\int\sin^2x\ \mathrm{dx}$ . In conclusion, $\boxed{\int \cos 2x\ln \cos x\ \mathrm{dx}=\frac 12\cdot \sin 2x\ln\cos x+\frac 12\cdot\left(x-\frac 12\cdot\sin 2x\right)+\mathbb C}$ .


PP2. Ascertain the definite integral $\int^{12}_0\frac{1}{\sqrt{4x+1}+\sqrt{x+4}}\ \mathrm{dx}$ .

Proof. $I=\frac 13\cdot \int^{12}_0\frac{\sqrt{4x+1}-\sqrt{x+4}}{x-1}\ \mathrm{dx}\implies$ $\boxed{\ I=\frac 13\cdot (U-V)\ }$ , where $U=\int^{12}_0\frac{\sqrt{4x+1}}{x-1}\ \mathrm{dx}$ and $V=\frac{\sqrt{x+4}}{x-1}\ \mathrm{dx}$ .

Method 1.

$\blacktriangleright\ U=\int^{12}_0\frac{\sqrt{4x+1}}{x-1}\ \mathrm{dx}\stackrel{(x=t^2+t)}{\ \ =\ \ }$ $\int_0^3\frac {(2t+1)^2}{t^2+t-1}\ \mathrm{dt}=$ $\int_0^3\left(4+\frac {5}{t^2+t-1}\right)\ \mathrm{dt}=$ $12+5\cdot \int_0^3\frac {\left(t+\frac 12\right)'}{\left(t+\frac 12\right)^2-\frac 54}\ \mathrm{dt}=$

$12+\sqrt 5\cdot\left|\ln\frac {2t+1-\sqrt 5}{2t+1+\sqrt 5}\right|_0^3=$ $12+\sqrt 5\cdot\ln\frac {\left(7-\sqrt 5\right)\left(\sqrt 5+1\right)}{\left(7+\sqrt 5\right)\left(\sqrt 5-1\right)}\implies$ $\boxed{U=12+2\sqrt 5\cdot\ln\frac {1+3\sqrt 5}{2}-\sqrt 5\cdot\ln 11}$ .

$\blacktriangleright\ V=\int^{12}_0\frac{\sqrt{x+4}}{x-1}\ \mathrm{dx}\stackrel{(x=t^2+4t)}{\ \ =\ \ }$ $\int_0^2\frac {2(t+2)^2}{t^2+4t-1}\ \mathrm{dt}=$ $2\cdot \int_0^2\left(1+\frac {5}{t^2+4t-1}\right)\ \mathrm{dt}=$ $4+10\cdot \int_0^2\frac {\left(t+2\right)'}{\left(t+2\right)^2-5}\ \mathrm{dt}=$

$4+\sqrt 5\cdot\left|\ln\frac {t+2-\sqrt 5}{t+2+\sqrt 5}\right|_0^2=$ $4+\sqrt 5\cdot\ln\frac {\left(4-\sqrt 5\right)\left(\sqrt 5+2\right)}{\left(4+\sqrt 5\right)\left(\sqrt 5-2\right)}\implies$ $\boxed{V=4+2\sqrt 5\cdot\ln (3+2\sqrt 5)-\sqrt 5\cdot\ln 11}$ .

Method 2.

$\blacktriangleright\ U=\int^{12}_0\frac{\sqrt{4x+1}}{x-1}\ \mathrm{dx}\stackrel{(4x+1=t^2)}{\ \ =\ \ }$ $2\cdot \int_1^7\frac {t^2}{t^2-5}\ \mathrm{dt}=$ $2\cdot \int_1^7\left(1+\frac {5}{t^2-5}\right)\ \mathrm{dt}=$ $12+\sqrt 5\cdot\left|\ln\frac {t-\sqrt 5}{t+\sqrt 5}\right|_1^7=$

$12+\sqrt 5\cdot\ln\frac {\left(7-\sqrt 5\right)\left(\sqrt 5+1\right)}{\left(7+\sqrt 5\right)\left(\sqrt 5-1\right)}\implies$ $\boxed{U=12+2\sqrt 5\cdot\ln\frac {1+3\sqrt 5}{2}-\sqrt 5\cdot\ln 11}$ .

$\blacktriangleright\ V=\int^{12}_0\frac{\sqrt{x+4}}{x-1}\ \mathrm{dx}\stackrel{(x+4=t^2)}{\ \ =\ \ }$ $2\cdot \int_2^4\frac {t^2}{t^2-5}\ \mathrm{dt}=$ $2\cdot \int_2^4\left(1+\frac {5}{t^2-5}\right)\ \mathrm{dt}=$ $4+\sqrt 5\cdot\left|\ln\frac {t-\sqrt 5}{t+\sqrt 5}\right|_2^4=$

$4+\sqrt 5\cdot\ln\frac {\left(4-\sqrt 5\right)\left(\sqrt 5+2\right)}{\left(4+\sqrt 5\right)\left(\sqrt 5-2\right)}\implies$ $\boxed{V=4+2\sqrt 5\cdot\ln (3+2\sqrt 5)-\sqrt 5\cdot\ln 11}$ .

In conclusion, $I=\frac 13\cdot\left[8+2\sqrt 5\cdot \ln\frac {1+3\sqrt 5}{2\left(3+2\sqrt 5\right)}\right]\implies$ $\boxed{I=\frac 83+\frac {2\sqrt 5}{3}\cdot\left(2\cdot\ln\frac {7-\sqrt 5}{2}-\ln 11\right)}$ .
This post has been edited 12 times. Last edited by Virgil Nicula, Nov 18, 2015, 5:55 AM

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53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by tigerzhang, Aug 1, 2021, 12:02 AM

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    by OlympusHero, Dec 30, 2020, 6:08 PM

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    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

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    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
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    by OlympusHero, May 14, 2020, 8:43 PM

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About Owner
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  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
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