أتحقق من فهمي

التكامل بالتعويض

التكاملات بالتعويض للتكاملات غير المحدودة

أتحقق من فهمي من فهمي صفحة (32):

أجد كلاً من التكاملات الآتية:

$\int 4x^2\sqrt{x^3-5}dx$ (a)

$\begin{array}{rl}u=x^3-5\Rightarrow \frac{du}{dx}& =3x^2\Rightarrow dx=\frac{du}{3x^2}\\ \int 4x^2\sqrt{x^3-5}dx=& \int 4x^2\sqrt{u}\times \frac{du}{3x^2}\\ & =\int \frac{4}{3}{u}^{\frac{1}{2}}du\\ & =\frac{8}{9}{u}^{\frac{3}{2}}+C\\ & =\frac{8}{9}\sqrt{{(x^3-5)}^3}+C\end{array}$

$\int \frac{1}{2\sqrt{x}}{e}^{\sqrt{x}}dx$ (b)

$\begin{array}{rl}& u=\sqrt{x}\Rightarrow \frac{du}{dx}=\frac{1}{2\sqrt{x}}\Rightarrow dx=2\sqrt{x}du\\ & \int \frac{1}{2\sqrt{x}}{e}^{\sqrt{x}}dx=\int \frac{1}{2\sqrt{x}}{e}^u\times 2\sqrt{x}du\\ & =\int e^{u}du\\ & =e^u+C\\ & =e^{\sqrt{x}}+C\end{array}$

$\int \frac{(\ln x{)}^3}{x}dx$ (c)

$\begin{array}{rl}& u=\ln x\Rightarrow \frac{du}{dx}=\frac{1}{x}\Rightarrow dx=xdu\\ & \begin{array}{rl}\int \frac{(\ln x{)}^3}{x}dx=\int & \frac{u^3}{x}\times xdu\\ & =\int u^{3}du\\ & =\frac{1}{4}{u}^4+C\\ & =\frac{1}{4}(\ln x{)}^4+C\end{array}\end{array}$

$\int \frac{\cos (\ln x)}{x}dx$ (d)

$\begin{array}{rl}u=\ln x\Rightarrow \frac{du}{dx}& =\frac{1}{x}\Rightarrow dx=xdu\\ \int \frac{\cos (\ln x)}{x}dx& =\int \frac{\cos u}{x}\times xdu\\ & =\int \cos udu\\ & =\sin u+C\\ & =\sin (\ln x)+C\end{array}$

$\int {\cos }^{4}5x\sin 5xdx$ (e)

$\begin{array}{rl}u=\cos 5x\Rightarrow \frac{du}{dx}& =-5\sin 5x\Rightarrow dx=\frac{du}{-5\sin 5x}\\ \int {\cos }^{4}5x\sin 5xdx& =\int u^4\sin 5x\times \frac{du}{-5\sin 5x}\\ & =\int -\frac{1}{5}{u}^{4}du\\ & =-\frac{1}{25}{u}^5+C\\ & =-\frac{1}{25}{\cos }^{5}5x+C\end{array}$

$\int x{2}^{x^2}dx$ (f)

$\begin{array}{rl}u=x^2\Rightarrow \frac{du}{dx}=2x& \Rightarrow dx=\frac{du}{2x}\\ \int x{2}^{x^2}dx=\int x{2}^u& \times \frac{du}{2x}\\ & =\int \frac{1}{2}{2}^{u}du\\ & =\frac{1}{2}\frac{2^u}{\ln 2}+C\\ & =\frac{1}{\ln 2}{2}^{x^2-1}+C\end{array}$

أتحقق من فهمي من فهمي صفحة (34):

أجد كلاً من التكاملات الآتية:

$\int \frac{x}{\sqrt{1+2x}}dx$ (a)

$\begin{array}{rl}& u=1+2x\Rightarrow \frac{du}{dx}=2\Rightarrow dx=\frac{du}{2},x=\frac{u-1}{2}\\ & \int \frac{x}{\sqrt{1+2x}}dx=\int \frac{\frac{1}{2}(u-1)}{u^{\frac{1}{2}}}\times \frac{du}{2}\\ & =\frac{1}{4}\int (u^{\frac{1}{2}}-u^{-\frac{1}{2}})du\\ & =\frac{1}{4}(\frac{2}{3}{u}^{\frac{3}{2}}-2u^{\frac{1}{2}})+C\\ & =\frac{1}{6}(1+2x{)}^{\frac{3}{2}}-\frac{1}{2}(1+2x{)}^{\frac{1}{2}}+C\\ & =\frac{1}{6}\sqrt{(1+2x{)}^3}-\frac{1}{2}\sqrt{1+2x}+C\end{array}$

$\int x^7{(x^4-8)}^{3}dx$ (b)

$\begin{array}{rl}u=x^4-8\Rightarrow \frac{du}{dx}& =4x^3\Rightarrow dx=\frac{du}{4x^3},x^4=u+8\\ \int x^7{(x^4-8)}^{3}dx& =\int x^7{u}^3\times \frac{du}{4x^3}\\ & =\frac{1}{4}\int x^4{u}^{3}du\\ & =\frac{1}{4}\int (u+8)u^{3}du\\ & =\frac{1}{4}\int (u^4+8u^3)du\\ & =\frac{1}{4}(\frac{1}{5}{u}^5+2u^4)+C\\ & =\frac{1}{20}{(x^4-8)}^5+\frac{1}{2}{(x^4-8)}^4+C\end{array}$

$\int \frac{e^{3x}}{{(1-e^x)}^2}dx$ (c)

$\begin{array}{rl}& u=1-e^x\Rightarrow \frac{du}{dx}=-e^x\Rightarrow dx=\frac{du}{-e^x},e^x=1-u\\ & \int \frac{e^{3x}}{{(1-e^x)}^2}dx=\int \frac{e^{3x}}{u^2}\times \frac{du}{-e^x}\\ & =\int -\frac{e^{2x}}{u^2}du\\ & =\int \frac{-(1-u{)}^2}{u^2}du\\ & =\int \frac{-1+2u-u^2}{u^2}du\\ & =\int (-u^{-2}+\frac{2}{u}-1)du\\ & =(u^{-1}+2\ln |u|-u)+C\\ & =\frac{1}{1-e^x}+2\ln |1-e^x|-1+e^x+C\end{array}$

التكاملات بالتعويض للتكاملات تحوي المقدار $\sqrt[n]{ax+b}$

أتحقق من فهمي من فهمي صفحة (35):

أجد كلاً من التكاملين الآتيين:

$\int \frac{dx}{x+\sqrt[3]{x}}$ (a)

$\begin{array}{rl}u=\sqrt[3]{x}& \Rightarrow \frac{du}{dx}=\frac{1}{3}{x}^{-\frac{2}{3}}\Rightarrow dx=3x^{\frac{2}{3}}du,x=u^3\\ \int \frac{dx}{x+\sqrt[3]{x}}& =\int \frac{3x^{\frac{2}{3}}du}{u^3+u}\\ & =\int \frac{3u^2}{u^3+u}du\\ & =\int \frac{3u}{u^2+1}du\\ & =\frac{3}{2}\int \frac{2u}{u^2+1}du\\ & =\frac{3}{2}\ln (u^2+1)+C\\ & =\frac{3}{2}\ln (x^{\frac{2}{3}}+1)+C\end{array}$

$\int x\sqrt[3]{(1-x{)}^2}dx$ (b)

$\begin{array}{rl}u=1-x\Rightarrow \frac{du}{dx}& =-1\Rightarrow dx=-du,x=1-u\\ \int x\sqrt[3]{(1-x{)}^2}dx& =\int x\sqrt[3]{u^2}\times -du\\ & =\int -(1-u{)}^{\sqrt[3]{u{u}^2}}du\\ & =\int -(1-u)u^{\frac{2}{3}}du\\ & =\int (-u^{\frac{2}{3}}+u^{\frac{5}{3}})du\\ & =-\frac{3}{5}{u}^{\frac{5}{3}}+\frac{3}{8}{u}^{\frac{8}{3}}+C\\ & =-\frac{3}{5}(1-x{)}^{\frac{5}{3}}+\frac{3}{8}(1-x{)}^{\frac{8}{3}}+C\\ & =-\frac{3}{5}\sqrt[3]{(1-x{)}^5}+\frac{3}{8}\sqrt[3]{(1-x{)}^8}+C\end{array}$

أتحقق من فهمي من فهمي صفحة (37):

أسعار: يمثل الاقتران $p\left(x\right)$ سعر قطعة (بالدينار) تستعمل في أجهزة الحاسوب، حيث $x$ عدد القطع المبيعة منها بالمئات. إذا كان: $p^{\mathrm{\prime }}\left(x\right)=\frac{-135x}{\sqrt{9+x^2}}$ هو معدل تغير سعر هذه القطعة، فأجد $p\left(x\right)$، علماً بأن سعر القطعة الواحدة هـو $30 \mathrm{JD}$ عندما يكون عدد القطع المبيعة منها 400 قطعة.

$\begin{array}{rl}& p\left(x\right)=\int \frac{-135x}{\sqrt{9+x^2}}dx\\ & u=9+x^2\Rightarrow \frac{du}{dx}=2x\Rightarrow dx=\frac{du}{2x}\\ & p\left(x\right)=\int \frac{-135x}{\sqrt{u}}\times \frac{du}{2x}\\ & =\frac{-135}{2}\int u^{-\frac{1}{2}}du\\ & =-135u^{\frac{1}{2}}+C\\ & p\left(x\right)=-135\sqrt{9+x^2}+C\\ & p\left(4\right)=-135\sqrt{9+16}+C=-135\left(5\right)+C\\ & 30=-675+C\Rightarrow C=705\\ & p\left(x\right)=705-135\sqrt{9+x^2}\end{array}$

التكامل بالتعويض لاقترانات تتضمن اقتراني الجيب وجيب التمام المرفوعين إلى أس فردي

أتحقق من فهمي من فهمي صفحة (39):

أجد كلاً من التكاملين الآتيين:

$\int {\sin }^{3}xdx$ (a)

$\begin{array}{rl}\int {\sin }^{3}xdx& =\int \sin x{\sin }^{2}xdx=\int \sin x(1-{\cos }^{2}x)dx\\ u=\cos x& \Rightarrow \frac{du}{dx}=-\sin x\Rightarrow dx=\frac{du}{-\sin x}\\ \int {\sin }^{3}xdx& =\int \sin x(1-u^2)\frac{du}{-\sin x}\\ & =\int (u^2-1)du\\ & =\frac{1}{3}{u}^3-u+C\\ & =\frac{1}{3}{\cos }^{3}x-\cos x+C\end{array}$

$\int {\cos }^{5}x{\sin }^{2}xdx$ (b)

$\begin{array}{rl}u=\sin x\Rightarrow \frac{du}{dx}& =\cos x\Rightarrow dx=\frac{du}{\cos x}\\ \int {\cos }^{5}x{\sin }^{2}xdx& =\int {\cos }^{5}x{u}^2\frac{du}{\cos x}\\ & =\int {\cos }^{4}x{u}^{2}du\\ & =\int {(1-{\sin }^{2}x)}^2{u}^{2}du\\ & =\int {(1-u^2)}^2{u}^{2}du\\ & =\int (u^2-2u^4+u^6)du\\ & =\frac{1}{3}{u}^3-\frac{2}{5}{u}^5+\frac{1}{7}{u}^7+C\\ & =\frac{1}{3}{\sin }^{3}x-\frac{2}{5}{\sin }^{5}x+\frac{1}{7}{\sin }^{7}x+C\end{array}$

التكامل بالتعويض لاقترانات تتضمن الظل أو ظل التمام، أو القاطع، أو قاطع التمام

أتحقق من فهمي من فهمي صفحة (41):

أجد كلاً من التكاملات الآتية:

$\int {\tan }^{4}xdx$ (a)

$\begin{array}{rl}\int {\tan }^{4}xdx& =\int {\tan }^{2}x{\tan }^{2}xdx\\ & =\int {\tan }^{2}x({\sec }^{2}x-1)dx\\ & =\int {\tan }^{2}x{\sec }^{2}x-{\tan }^{2}xdx\\ & =\int {\tan }^{2}x{\sec }^{2}xdx-\int {\tan }^{2}xdx\\ & =\int {\tan }^{2}x{\sec }^{2}xdx-\int ({\sec }^{2}x-1)dx\\ u=\tan x& \Rightarrow \frac{du}{dx}={\sec }^{2}x\Rightarrow dx=\frac{du}{{\sec }^{2}x}\\ \Rightarrow \tan x& dx=\int u^2{\sec }^{2}x\times \frac{du}{{\sec }^{2}x}-\int ({\sec }^{2}x-1)dx\\ & =\int u^{2}du-\int ({\sec }^{2}x-1)dx\\ & =\frac{1}{3}{u}^3-\tan x+x+C\\ & =\frac{1}{3}{\tan }^{3}x-\tan x+x+C\end{array}$

$\int {\cot }^{5}xdx$ (b)

$\begin{array}{rl}& \begin{array}{rl}\int {\cot }^{5}xdx& =\int \cot x{\cot }^{4}xdx\\ & =\int \cot x{({\cot }^{2}x)}^{2}dx\\ & =\int \cot x{({\csc }^{2}x-1)}^{2}dx\\ u=\csc x\Rightarrow & \frac{du}{dx}=-\csc x\cot x\Rightarrow dx=\frac{du}{-\csc x\cot x}\\ \Rightarrow \int {\cot }^{5}xdx& =\int \cot x{(u^2-1)}^2\times \frac{du}{-\csc x\cot x}\\ & =\int {(u^2-1)}^2\frac{du}{-u}\\ & =\int \frac{u^4-2u^2+1}{-u}du\\ & =\int (-u^3+2u-\frac{1}{u})du\\ & =-\frac{1}{4}{u}^4+u^2-\ln \left|u\right|+C\\ & =-\frac{1}{4}{\csc }^{4}x+{\csc }^{2}x-\ln |\csc x|+C\end{array}\end{array}$

$\int {\sec }^{4}x{\tan }^{6}xdx$ (c)

$\begin{array}{rl}u=\tan x\Rightarrow \frac{du}{dx}& ={\sec }^{2}x\Rightarrow dx=\frac{du}{{\sec }^{2}x}\\ \int {\sec }^{4}x{\tan }^{6}xdx& =\int {\sec }^{4}x{u}^6\times \frac{du}{{\sec }^{2}x}\\ & =\int {\sec }^{2}x{u}^{6}du\\ & =\int (1+{\tan }^{2}x)u^{6}du\\ & =\int (1+u^2)u^{6}du\\ & =\int (u^6+u^8)du\\ & =\frac{1}{7}{u}^7+\frac{1}{9}{u}^9+C\\ & =\frac{1}{7}{\tan }^{7}x+\frac{1}{9}{\tan }^{9}x+C\end{array}$

التكامل بالتعويض للتكاملات المحدودة

أتحقق من فهمي من فهمي صفحة (43):

أجد قيمة كل من التكاملين الآتيين:

${\int }_0^{2}x(x+1{)}^{3}dx$ (a)

$\begin{array}{rl}& u=x+1\Rightarrow \frac{u}{dx}=1\Rightarrow dx=du,x=u-1\\ & x=0\Rightarrow u=1\\ & x=2\Rightarrow u=3\\ & {\int }_0^{2}x(x+1{)}^{3}dx={\int }_1^3(u-1)u^{3}du\\ & ={\int }_1^3(u^4-u^3)du\\ & ={(\frac{1}{5}{u}^5-\frac{1}{4}{u}^4)|}_1^3\\ & =\frac{1}{5}(3{)}^5-\frac{1}{4}(3{)}^4-\left(\frac{1}{5}\right(1{)}^5-\frac{1}{4}\left(1{)}^4\right)\\ & =\frac{142}{5}=28.4\\ & \end{array}$

${\int }_0^{\pi /3}\sec x\tan x\sqrt{\sec x+2}dx$ (b)

$\begin{array}{rl}& u=\sec x+2\Rightarrow \frac{du}{dx}=\sec x\tan x\Rightarrow dx=\frac{du}{\sec x\tan x}\\ & x=0\Rightarrow u=3\\ & x=\frac{\pi }{3}⟹u=4\\ & {\int }_0^{\frac{\pi }{3}}\sec x\tan x\sqrt{\sec x+2}dx={\int }_3^4\sec x\tan x\sqrt{u}\frac{du}{\sec x\tan x}\\ & ={\int }_3^4\sqrt{u}du\\ & ={\frac{2}{3}{u}^{\frac{3}{2}}|}_3^4\\ & =\frac{2}{3}(8-3\sqrt{3})\approx 1.87\end{array}$