\[\int \frac{2}{1\dotplus \sin x-cos x }\]
عرض للطباعة
\[\int \frac{2}{1\dotplus \sin x-cos x }\]
$\int \frac{2}{1+sinx-cosx}dx$
$=\int \frac{2sin^{2}\frac{x}{2}+2cos^{2}\frac{x}{2}}{sin ^{2}\frac{x}{2}+cos^{2}\frac{x}{2}+2sin\frac{x}{2} cos\frac{x}{2}-cos^{2}\frac{x}{2}+sin^{2}\frac{x}{2}}dx$
$=\int \frac{2sin\frac{x}{2}cos\frac{x}{2}+2cos^{2}\frac{ x}{2}-2sin\frac{x}{2}cos\frac{x}{2}+2sin^{2}\frac{x}{2}} {2sin^{2}\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2} }dx$
$=\int \frac{2sin\frac{x}{2}cos\frac{x}{2}+2cos^{2}\frac{ x}{2}}{2sin^{2}\frac{x}{2}+2sin\frac{x}{2}cos\frac {x}{2}}dx-\int \frac{2sin\frac{x}{2}cos\frac{x}{2}+2sin^{2}\frac{ x}{2}}{2sin^{2}\frac{x}{2}+2sin\frac{x}{2}cos\frac {x}{2}}dx$
$=\int \frac{cos\frac{x}{2}}{sin\frac{x}{2}}dx-\int \frac{cos\frac{x}{2}-sin\frac{x}{2}}{sin\frac{x}{2}+cos\frac{x}{2}}dx$
$=2ln\left | sin\frac{x}{2} \right |-2ln\left | sin\frac{x}{2}+cos\frac{x}{2} \right |+c$
\[\int \frac{2}{1+sinx-cosx}{\color{Red} \times \frac{1-\left ( sinx-cosx \right )}{1-\left ( sinx-cosx \right )}}dx=2\int\frac{1-sinx+cosx}{2sinxcosx} dx\\=\int \left [ 2csc2x {\color{Red} \times \frac{csc2x+cot2x}{csc2x+cot2x}}+secx{\color{Red} \times \frac{secx+tanx}{secx+tanx}}+cscx {\color{Red} \times \frac{cscx+cotx}{cscx+cotx}}\right ]dx \\=-ln|csc2x+cot2x|+ln|secx+tanx|-ln|cscx+cotx|+C\]اقتباس:
المشاركة الأصلية كتبت بواسطة مدرس مادة الرياضيات
حل اخر
$\int \frac{2}{1+sinx-cosx}dx$
$=\int \frac{2}{1+\frac{2tan\frac{x}{2}}{1+tan^{2}\frac{x }{2}}-\frac{1-tan^{2}\frac{x}{2}}{1+tan^{2}\frac{x}{2}}}dx$
$=\int \frac{1+tan^{2}\frac{x}{2}}{tan\frac{x}{2}+tan^{2} \frac{x}{2}}dx$
$=\int \frac{sec^{2}\frac{x}{2}}{tan\frac{x}{2}(1+tan \frac{x}{2})}dx$
$=\int \frac{sec^{2}\frac{x}{2}}{tan\frac{x}{2}}dx-\int \frac{sec^{2}\frac{x}{2}}{(1+tan\frac{x}{2})}dx$
$=2ln\left | tan\frac{x}{2} \right |-2ln\left | 1+tan\frac{x}{2} \right |+c$
$=2ln\left | \frac{tan\frac{x}{2}}{1+tan\frac{x}{2}} \right |+c$