سؤال تكامل للحل ( تجربة طباعة باللاتكس)

$\int_{1}^{5}\frac{x}{\sqrt{2x-1}}dx$

$=\int_{1}^{5}\frac{1+2x-1}{2\sqrt{2x-1}}dx$

$=\int_{1}^{5}\frac{1}{2\sqrt{2x-1}}dx+\int_{1}^{5}\frac{1}{2}\sqrt{2x-1}dx$

$=\frac{1}{2}\sqrt{2x-1}+\frac{1}{6}(2x-1)^{\frac{3}{2}}]_{1}^{5}$

$=\frac{16}{3}$

السلام عليكم ورحمة الله
مجرد تجربة العمل بنظام الطباعة الجديد[IMG]\[$I=\int_{1}^{2}{x(\sqrt{2x-1})}dx , put, 2x-1=u\rightarrow dx=1/2du I=1/4\int_{1}^{3}{u^{1/2}(u+1)}du=1/4\int_{1}^{3}(u^{3/2}+u^{1/2})du\][/IMG]

\[\LARGE Q5.\int_{0}^{1} \arccos x\: dx\]

$\int_{0}^{1}\arccos x dx$

$=\int_{0}^{1}(arccosx-\frac{x}{\sqrt{1-x^{2}}}+\frac{x}{\sqrt{1-x^{2}}})dx$

$=\int_{0}^{1}(arccosx-\frac{x}{\sqrt{1-x^{2}}})dx-\int_{0}^{1}(\frac{-2x}{2\sqrt{1-x^{2}}})dx$

$=xaccosx-\sqrt{1-x^{2}}]_{0}^{1}=1$

حل اخر
$\int_{0}^{1}\arccos x dx$

$u=arccosx\Rightarrow du=\frac{-1}{\sqrt{1-x^{2}}}dx,v=x$

$=xaccosx]_{0}^{1}+\int_{0}^{1}(\frac{x}{\sqrt{1-x^{2}}})dx=0+(-\sqrt{1-x^{2}}]_{0}^{1})=1$

\[\LARGE Q6.\int_{0}^{1} \arctan x\: dx\]

$\int_{0}^{1}\arctan xdx$

$=\int_{0}^{1}(arctan x+\frac{x}{1+x^{2}}-\frac{x}{1+x^{2}})dx$

$=\int_{0}^{1}((arctan x+\frac{x}{1+x^{2}})-\frac{1}{2}\frac{2x}{1+x^{2}})dx$

$=xactanx-\frac{1}{2}ln(x^{2}+1)]_{0}^{1}=\frac{\pi }{4}-\frac{1}{2}ln2$

\[\LARGE Q7.\int_{0}^{1}\arccos \left ( \frac{1-x^2}{1+x^2} \right )\: dx\]

$\int_{0}^{1}arccos(\frac{1-x^{2}}{1+x^{2}})dx$

$x=tany\rightarrow dx=sec^{2}ydy$

$=\int_{0}^{\frac{\pi }{4}}arccos(\frac{1-tan^{2}y}{1+tan^{2}y})sec^{2}ydy$

$=\int_{0}^{\frac{\pi }{4}}arccos(cos2y)sec^{2}ydy=\int_{0}^{\frac{\pi }{4}}2ysec^{2}ydy$

$=2y tany]_{0}^{\frac{\pi }{4}}-2\int_{0}^{\frac{\pi }{4}}tanydy$

$=2y tany+2ln\left | cosy \right |]_{0}^{\frac{\pi }{4}}=\frac{\pi }{2}+2ln\frac{1}{\sqrt{2}}$

\[\LARGE Q8.\int_{0}^{1}\arctan \left ( \frac{2x}{1-x^2} \right )\: dx\]

$\int_{0}^{1}arctan(\frac{2x}{1-x^{2}})dx$

$x=tany\rightarrow dx=sec^{2}ydy$

$=\int_{0}^{\frac{\pi }{4}}arctan(\frac{2 tany}{1-tan^{2}y})sec^{2}ydy$

$=\int_{0}^{\frac{\pi }{4}}arctan(tan2y)sec^{2}ydy=\int_{0}^{\frac{\pi }{4}}2ysec^{2}ydy$

$=2y tany]_{0}^{\frac{\pi }{4}}-2\int_{0}^{\frac{\pi }{4}}tanydy$

$=2y tany+2ln\left | cosy \right |]_{0}^{\frac{\pi }{4}}=\frac{\pi }{2}+2ln\frac{1}{\sqrt{2}}$