عرض للطباعة
$Q1)\int \frac{\sqrt[3]{x-1}+\sqrt[4]{x-1}}{(x-1)(1+\sqrt[6]{x-1})}dx$
$Q2)\int (1-\frac{1}{x^{2}})\sqrt{x\sqrt{x}}dx$
$Q3)\int \frac{\sqrt{4-x^{2}}}{x^{2}}dx$
$Q4)\int \frac{cscx}{ln(tan\frac{x}{2})}dx$
$Q5)\int \frac{sin(lnx)}{x^{3}}dx$
$Q6)\int \frac{cos2x-1}{cos2x+1}dx$
$Q7)\int \frac{sinx-cosx}{\sqrt{2sinxcosx}}dx$
$Q8)\int \frac{1}{sinxcos^{2}x}dx$
$Q9)\int \sqrt{\frac{1-cosx}{cosx(1+cosx)(2+cosx)}}dx$
$Q10)\int \frac{1}{sin^{4}xcos^{2}x}dx$
بسم الله . جواب Q4
http://alnasiry.net/forums/uploaded/16811_mas155.gif
بسم الله . لاعطاء جمالية اكثر للسؤال الرابع Q4 , هذا حل ثاني للسؤال Q4 الجميل للاستاذ فلاح الناصري المحترم :
http://alnasiry.net/forums/uploaded/16811_mas149.gif
\[{\color{Red} Q1})y^{12}=x-1\Rightarrow 12y^{11}dy=dx\\\int \frac{\sqrt[3]{x-1}+\sqrt[4]{x-1}}{(x-1)(1+\sqrt[6]{x-1})}=12\int \frac{y^{4}+{\color{Red} y^{2}}+y^{3}-{\color{Red} y^{2}}}{y\left ( 1+y^{2} \right )}dy\\=12\int ydy+12\int \frac{y^{2}-y}{y^{2}+1}dy=6y^{2}+12\int \frac{y^{2}+1-y-1}{y^{2}+1}dy\\=6y^{2}+12\int dy-12\int \frac{y}{y^{2}+1}dy-12\int \frac{1}{y^{2}+1}dy\\=6y^{2}+12y-6ln|y^{2}+1|-12tan^{-1}y+C\\=6\sqrt[6]{x-1}+12\sqrt[12]{x-1}-6ln|\sqrt[6]{x-1}+1|-12tan^{-1}\sqrt[6]{x-1}+C\]
\[Q2)x=y^{4}\Rightarrow dx=4y^{3}dy\\\int \left ( 1-\frac{1}{x^{2}} \right )dx=\int \left ( 1-\frac{1}{x^{2}} \right )\sqrt{x\sqrt{x}}dx\\=4\int \left ( 1-\frac{1}{y^{8}} \right ).y^{6}dy=4\int \left ( y^{6}-y^{-2} \right )dy\\=\frac{4}{7}y^{7}+4y^{-1}+C=\frac{4}{7}x^{\frac{7}{4}}+\frac{4}{\sqrt[4]{x}}+C\]
بسم الله . جواب السؤال الخامس Q5
http://alnasiry.net/forums/uploaded/16811_mas151.gif
بسم الله . هذا حل ثاني للسؤال الخامس Q5
http://alnasiry.net/forums/uploaded/16811_ma150.gif
$Q11)\int \frac{1}{\sqrt{sin^{3}x sin(x+3)}}dx$
$Q12)\int \frac{cos5x+cos4x}{1-2cos3x}dx$
$Q13)\int \frac{1}{1+2sin^{2}x}dx$
$Q14) \int \frac{tanx}{cotx+tanx}dx$
$Q15)\int \frac{sinx}{cotx+tanx}dx$
$Q16)\int \frac{cosx}{cotx+tanx}dx$
$Q17)\int \frac{cosx}{2+3sinx+sin^{2}x}dx$
$Q18)\int \frac{cosx}{1+2sinx+sin^{2}x}dx$
$Q19)\int \sqrt{cotx}dx$
$Q20)\int (\sqrt{cotx}+\sqrt{tanx})dx$
بسم الله . نظرا لجمالية السؤال Q10 , هذا حل ثاني بعد الحل الجميل للاستاذ جليل الزيادي المحترم .
http://alnasiry.net/forums/uploaded/16811_mas152.gif
[IMG]$\huge {\color{Red} \int (\frac{\tan x}{\cot x+\tan x})dx=\int (\frac{1}{\cot ^{2}x+1})dx =\int \sin ^{2}xdx=1/2[x-1/2\sin 2x]+c}$ [/IMG]
$\huge {\color{Blue} \int (\frac{\sin }{\cot x+\tan x})dx=\int (\frac{\sin x}{\frac{1}{\sin x\cos x}})dx=\int(\sin ^{2}x\cos x)dx=1/3 \sin ^{3}x+c }$
$\huge {\color{Blue}\int (\frac{\cos x}{\tan x+\cos x})dx=\int (\frac{\cos x}{\frac{1}{\sin x\cos x}})dx=\int (\cos ^{2}x)\sin xdx=-1/3\cos ^{3}x+c }$
سؤال12 للتذكير ؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟ ؟؟؟؟؟
بسم الله . حل سؤال 12 / Q12 تبدو الصورة كبيرة علما ان حجمها 52.1 kB
http://alnasiry.net/forums/uploaded/16811_mas154.gif
$Q21)\int \frac{cos^{3}x+cos^{5}x}{sin^{2}x+sin^{4}x}dx$
$Q22)\int \frac{tanx+tan^{3}x}{1+tan^{3}x}dx$
$Q23)\int \frac{sinx}{sin4x}dx$
$Q24)\int \frac{1}{\sqrt{sin^{5}xcos^{3}x}}dx$
$Q25)\int \frac{1}{sin^{3}x cos^{5}x}dx$
$Q26)\int \frac{x}{1+cosx}dx$
$Q27)\int \frac{sinhx}{sinhx+coshx}dx$
$Q28)\int sinx cosx(sin2x+cos2x)dx$
$Q29)\int \frac{cos2x-cos4}{cosx-cos2}dx$
$Q30)\int \frac{(cosx-sinx)(1+sin2x)}{sinx+cosx}dx$
بسم الله . لجمالية السؤال 25 / Q25 , هذا حله بطريقة ثانية :
http://alnasiry.net/forums/uploaded/16811_mas156.gif
$Q31)\int \frac{sec3x}{secx}dx$
$Q32)\int \frac{csc3x}{cscx}dx$
$Q33)\int \frac{arccotx}{x^{3}}dx$
$Q34)\int arccot(cscx+cotx)dx$
$Q35)\int tan(x-2)tan(x+2)tan2x dx$
\[Q31)\because sec3x=\frac{1}{cos3x}=\frac{1}{4cos^3x-3cosx}=\frac{sec^{3}x}{4-3sec^2x}\\\therefore \int \frac{sec^{3}x}{secx}dx=\int \frac{sec^{2}x}{4-3sec^{2}x}dx=\int \frac{sec^{2}x}{1-3tan^{2}x}dx \\=\frac{1}{\sqrt{3}}\tanh ^-1\left ( \sqrt{3}tanx \right )+C\]
\[Q34)\int arc cot\left ( cscx+cotx \right )dx=\int arccot\left ( \frac{1+cosx}{sinx} \right )dx\\=\int arccot\left ( \frac{2cos^{2}\frac{x}{2}}{2sin\frac{x}{2}cos\frac {x}{2}} \right )dx=\int arccot\left ( cot\frac{x}{2} \right )dx=\frac{x^{2}}{4}+C\]
[]$\huge {\color{Blue} \int(\tan(x+2)\tan(x-2)\tan2x)dx}$
$\huge {\color{Blue} \tan2x=\tan(x+2+x-2)=\frac{\tan(x-2)+\tan(x+2)}{1-\tan(x-2)\tan(x+2)}\rightarrow }$
$\huge {\color{Blue} \tan(x+2)\tan(x-2)=1-\frac{\tan(x+2)+\tan(x-2)}{\tan2x}}$
$\huge {\color{Blue} I=\int(\tan2x-\tan(x+2)+\tan(x-2))dx=-1/2 \ln\cos(2x)+\ln\cos(x+2)-\ln\cos(x-2)+c}$[/]