ÇáÓáÇã Úáíßã æÑÍãÉ Çááå æÈÑßÇÊå
\[\huge {\color{Red} {\color{Red} \int \frac{sinx}{cosx(1+cos^2x)}dx}}\]
ÇáÓáÇã Úáíßã æÑÍãÉ Çááå æÈÑßÇÊå
\[\huge {\color{Red} {\color{Red} \int \frac{sinx}{cosx(1+cos^2x)}dx}}\]
ÇáÓáÇã Úáíßã æÑÍãÉ Çááå æÈÑßÇÊå íÇãæÌåäÇ ÇáßÈíÑ ÇÈæ ÇÍãÏ
äÖÚ y=cosx æäÓÊÎÏã Çáß ÇáÌÒÆíå
ÇáÓáÇã Úáíßã æÑÍãÉ Çááå
$\LARGE {\color{DarkBlue} \int \frac{sinx}{cosx(1+cos^{2}x)}dx=\int \frac{tanx*sec^{2}x}{sec^{2}x+1}dx=\frac{1}{2}ln|s ec^{2}x+1|+c}$
$\int \frac{x^{2}}{1+x^{6}}dx$
$=\frac{1}{3}\int \frac{3x^{2}}{1+(x^{3})^{2}}dx$
$=\frac{1}{3}arctan(x^{3})+c$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
ÇáÓáÇã Úáíßã æÑÍãÉ Çááå æÈÑßÇÊå
\[\huge {\color{Red} {\color{Red} \int \frac{x-1}{(x+1)\sqrt{x(x^2+x+1)}}dx}}\]
íáÛì ÇáÍá áæÌæÏ ÇÎØÇÁ
ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ ÕáÇÍ ÇÍãÏ ; 06-30-2016 ÇáÓÇÚÉ 02:29 PM
ÇáÓáÇã Úáíßã æÑÍãÉ Çááå æÈÑßÇÊå
\[\huge {\color{Red} {\color{Red} \int_{0}^{\pi }ln\frac{1+xcos\theta }{1-xcos\theta }\frac{d\theta }{cos\theta }such that\left | x \right |< 1}}\]
ÇáÐíä íÔÇåÏæä ÇáãæÖæÚ ÇáÂä: 1 (0 ãä ÇáÃÚÖÇÁ æ 1 ÒÇÆÑ)
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