سلسلة التكامل

[]$\huge {\color{Blue} \int(\tan(x+2)\tan(x-2)\tan2x)dx}$
$\huge {\color{Blue} \tan2x=\tan(x+2+x-2)=\frac{\tan(x-2)+\tan(x+2)}{1-\tan(x-2)\tan(x+2)}\rightarrow }$
$\huge {\color{Blue} \tan(x+2)\tan(x-2)=1-\frac{\tan(x+2)+\tan(x-2)}{\tan2x}}$
$\huge {\color{Blue} I=\int(\tan2x-\tan(x+2)+\tan(x-2))dx=-1/2 \ln\cos(2x)+\ln\cos(x+2)-\ln\cos(x-2)+c}$[/]

$Q36)\int arccot(\frac{sinx+cosx}{cosx-sinx})dx$

$Q37)\int \frac{1-cotx}{1+cotx}dx$

$Q38)\int \frac{csc3x csc5x}{csc2x}dx$

$Q39)\int tanx tan(x+(\frac{\pi }{3}))tan(x-(\frac{\pi }{3}))dx$

$Q40)\int \frac{1}{csc^{2}x+cot^{2}x}dx$

$Q36)\int arccot(\frac{sinx+cosx}{cosx-sinx})dx$

$Q37)\int \frac{1-cotx}{1+cotx}dx$

$Q38)\int \frac{csc3x csc5x}{csc2x}dx$

$Q39)\int tanx tan(x+(\frac{\pi }{3}))tan(x-(\frac{\pi }{3}))dx$

$Q40)\int \frac{1}{csc^{2}x+cot^{2}x}dx$

بسم الله . حل سؤال 39 / Q39

بسم الله . حل سؤال 40 / Q40

$Q41)\int \frac{tan2-tanx}{tan2+tanx}dx$

$Q42)\int sin^{3}x cos\frac{x}{2}dx$

$Q43)\int \frac{secx}{\sqrt{sin(2x+2)+sin2}}dx$

$Q44)\int \frac{1}{\sqrt[3]{sin^{11}x cosx}}dx$

$Q45)\int \frac{cos^{3}x}{(sin^{4}x+3sin^{2}x+1)arccot(cscx+ sinx)}dx$

$\large {\color{Red} \int (\frac{\tan2x-tanx}{\tan2x+tanx})dx=\int (\frac{\frac{2tanx}{1-tan^{2}x}-tanx}{\frac{2tanx}{1-tan^{2}x}+tanx})dx=\int (\frac{2tanx+tan^{3}x-tanx}{2tanx-tan^{3}x+tan})dx}$
${\color{Red} =\int (\frac{sec^{2}x}{3-tan^{2}x})dx=\int (\frac{dy}{3-y^{2}})=\frac{1}{\sqrt{3}}\tanh^{-1}(1/\sqrt{3})y+c}$

$\large {\color{Blue} \int sin^{3}xcos(x/2))dx=8\int sin^{3}(x/2)cos^{4}(x/2)dx=8\int sin(x/2)cos^{4}(x/2)[1-cos^{2}(x/2)]dx}$
$ {\color{Blue} =8[\frac{-2}{a}cos^{a}(x/2)+\frac{2}{7}cos^{7}(x/2)]+c}$