تمارين (4-4)

(1)- جد تكاملات كل مما يلي ضمن مجال الدالة:

$\int \frac{{(2x^2-3)}^2-9}{x^2}dx$

$\begin{array}{rl}\int \frac{{(2x^2-3)}^2-9}{x^2}dx& =\int \frac{(4x^4-12x^2+9)-9}{x^2}dx=\int (\frac{4x^4}{x^2}-\frac{12x^2}{x^2})dx\\ & =\int (4x^2-12)dx=\frac{4}{3}{x}^3-12x+c\end{array}$

$\int \frac{(3-\sqrt{5x}{)}^7}{\sqrt{7x}}dx$

$$\begin{gathered} \sqrt{7x}=\sqrt{7}\sqrt{x} , \sqrt{5x}=\sqrt{5}\sqrt{x} \\ =\frac{1}{\sqrt{7}}\int \frac{(3-\sqrt{5}\sqrt{x}{)}^7}{x^{\frac{1}{2}}}=\frac{1}{\sqrt{7}}\int {(3-\sqrt{5}{x}^{\frac{1}{2}})}^7\cdot x^{-\frac{1}{2}} \\ \begin{array}{rl}& =\frac{-2}{\sqrt{5}}\frac{1}{\sqrt{7}}\int {(3-\sqrt{5}{x}^{\frac{1}{2}})}^7\left(\underbrace{-\frac{\sqrt{5}}{2}{x}^{-\frac{1}{2}}}\right)dx=\frac{-2}{\sqrt{35}}\frac{(3-\sqrt{5x}{)}^8}{8}+c\\ & =\frac{-1}{4\sqrt{35}}(3-\sqrt{5x}{)}^8+c\end{array} \end{gathered}$$

$\int \frac{{\cos }^{3}x}{1-\sin x}dx$

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

$\int {\csc }^{2}x\cos xdx$

$\begin{array}{rl}& \int {\csc }^{2}x\cos xdx\left(\csc x=\frac{1}{\sin x}\right)\\ & =\int \csc x\csc x\cos xdx=\int \csc x\frac{1}{\sin x}\cos xdx , \left(\frac{\cos x}{\sin x}=\cot \right)x\\ & =\int \csc x\cot xdx=-\csc x+c{\csc }^{2}x=\frac{1}{{\sin }^{2}x} \left(\mathrm{بوضع} \mathrm{آخر} \mathrm{بحل} \mathrm{ويحل}\right)\end{array}$

$\begin{array}{rl}& \int \frac{x}{{(3x^2+5)}^4}dx\end{array}$

$\begin{array}{rl}& \int \frac{x}{{(3x^2+5)}^4}dx=\int {(3x^2+5)}^{-4}\cdot xdx\\ & =\frac{1}{6}\int {(3x^2+5)}^{-4}\cdot \left(6x\right)dx=\frac{1}{6}\frac{{(3x^2+5)}^{-3}}{-3}+c=\frac{-1}{18{(3x^2+5)}^3}+c\end{array}$

$\int \sqrt[3]{x^2+10x+25}dx$

$\begin{array}{rl}& \int \sqrt[3]{x^2+10x+25}dx=\int \sqrt[3]{(x+5{)}^2}dx \left(\mathrm{كامل} \mathrm{مربع} \mathrm{المقدار} \mathrm{نجعل}\right)\\ & =\int (x+5{)}^{\frac{2}{3}}dx=\frac{(x+5{)}^{\frac{5}{3}}}{5}+c=\frac{3}{5}(x+5{)}^{\frac{5}{3}}+c \left(1= \mathrm{القوس} \mathrm{داخل} \mathrm{مشتقة}\right)\end{array}$

$\begin{array}{rl}& \int {\sin }^{3}xdx\end{array}$

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

$\int \frac{\cos \sqrt{1-x}}{\sqrt{1-x}}dx$

المقام هو مشتقة تكامل الـ $\cos$

$$\begin{gathered} =\int \frac{\cos (1-x{)}^{\frac{1}{2}}}{(1-x{)}^{\frac{1}{2}}}dx=-2\int \cos (1-x{)}^{\frac{1}{2}}\underset{\mathrm{الزاوية} \mathrm{مشتقة}}{\underbrace{(-\frac{1}{2}(1-x{)}^{-\frac{1}{2}})}}dx \\ =-2\sin (1-x{)}^{\frac{1}{2}}+c=-2\sin \sqrt{1-x}+c \end{gathered}$$

لو كان السؤال أعلاه بالشكل التالي $\int \frac{\sin \sqrt{1-x}}{\sqrt{1-x}}dx$

المقام هو مشتقة تكامل الـ $\sin$

$\begin{array}{rl}& =\int \frac{\sin (1-x{)}^{\frac{1}{2}}}{(1-x{)}^{\frac{1}{2}}}dx=-2\int \sin (1-x{)}^{\frac{1}{2}}\underset{\mathrm{الزاوية} \mathrm{مشتقة}}{\underbrace{(-\frac{1}{2}(1-x{)}^{-\frac{1}{2}})}}dx\\ & =2\cos (1-x{)}^{\frac{1}{2}}+c=2\cos \sqrt{1-x}+c\end{array}$

$\int {(3x^2+1)}^{2}dx$

مشتقة داخل القوس غير موجودة نفتح الأقواس.

$=\int (9x^4+6x^2+1)dx=\frac{9x^5}{5}+\frac{6x^3}{3}+x+c$

$\int \frac{\sqrt{\sqrt{x}-x}}{\sqrt[4]{x^3}}$

$\begin{array}{rl}& =\int \frac{\sqrt{\sqrt{x}-\sqrt{x}\sqrt{x}}}{x^{\frac{3}{4}}}=\int \frac{\sqrt{\sqrt{x}(1-\sqrt{x})}}{x^{\frac{3}{4}}}dx=\int \frac{{\left(x^{\frac{1}{2}}\right)}^{\frac{1}{2}}{(1-x^{\frac{1}{2}})}^{\frac{1}{2}}}{x^{\frac{3}{4}}}dx\\ & =\int {(1-x^{\frac{1}{2}})}^{\frac{1}{2}}{x}^{\frac{1}{4}}\cdot x^{\frac{-3}{4}}dx=\int {(1-x^{\frac{1}{2}})}^{\frac{1}{2}}\cdot x^{-\frac{1}{2}}dx\\ & =-2\int {(1-x^{\frac{1}{2}})}^{\frac{1}{2}}\cdot \underset{\mathrm{القوس} \mathrm{داخل} \mathrm{مشتقة}}{\underbrace{(-\frac{1}{2}{x}^{-\frac{1}{2}})}}dx=-2\frac{(1-x^{{\frac{1}{2})}^{\frac{3}{2}}}}{\frac{3}{2}}+c=-4\frac{{(1-x^{\frac{1}{2}})}^{\frac{3}{2}}}{3}+c\end{array}$

$x=\sqrt{x}\sqrt{x}$ لو كان السؤال أعلاه بالشكل التالي $\int \frac{\sqrt{x-\sqrt{x}}}{\sqrt[4]{x^3}}dx$

$\sqrt{x}$ نستخرجه عامل مشترك

$\begin{array}{rl}& =\int \frac{\sqrt{\sqrt{x}\sqrt{x}-\sqrt{x}}}{x^{\frac{3}{4}}}\\ & =\int \frac{\sqrt{\sqrt{x}(\sqrt{x}-1)}}{x^{\frac{3}{4}}}dx=\int \frac{{\left(x^{\frac{1}{2}}\right)}^{\frac{1}{2}}{(x^{\frac{1}{2}}-1)}^{\frac{1}{2}}}{x^{\frac{3}{x}}}dx\\ & =\int {(x^{\frac{1}{2}}-1)}^{\frac{1}{2}}\cdot x^{\frac{1}{4}}{x}^{\frac{-3}{4}}dx=\int {(x^{\frac{1}{2}}-1)}^{\frac{1}{2}}\cdot x^{-\frac{1}{2}}dx\\ & =2\int {(x^{\frac{1}{2}}-1)}^{\frac{1}{2}}\cdot \underset{\mathrm{القوس} \mathrm{داخل} \mathrm{مشتقة}}{\underbrace{\left(\frac{1}{2}{x}^{-\frac{1}{2}}\right)}}dx=2\frac{{(x^{\frac{1}{2}}-1)}^{\frac{3}{2}}}{\frac{3}{2}}+c=4\frac{{(x^{\frac{1}{2}}-1)}^{\frac{3}{2}}}{3}+c\end{array}$

$\int (1+\cos 3x{)}^{2}dx$

$$\begin{gathered} \begin{array}{rl}& \int (1+\cos 3x{)}^{2}dx\\ & =\int (1+2\cos 3x+{\cos }^{2}3x)dx\left({\cos }^{2}3x=\frac{1}{2}(1+\cos 6x)\right)\end{array} \\ \begin{array}{rl}& =\int (1+2\cos 3x+\frac{1}{2}(1+\cos 6x\left)\right)dx\\ & =\int dx+\frac{1}{3}\int 2\cos 3x\left(3\right)dx+\frac{1}{2}[\int dx+\frac{1}{6}\int \cos 6x(6\left)dx\right]\\ & =[x+2\frac{\sin 3x}{3}+\frac{1}{2}(x+\frac{\sin 6x}{6})]+c=\frac{3}{2}x+\frac{2}{3}\sin 3x+\frac{1}{12}\sin 6x+c\end{array} \end{gathered}$$

$\int {\sec }^{2}4xdx$

$\int {\sec }^{2}4xdx=\frac{1}{4}\int {\sec }^{2}4x\left(4\right)dx=\frac{\tan 4x}{4}+c$

$\int {\csc }^{2}2xdx$

$\int {\csc }^{2}2xdx=\frac{1}{2}\int {\csc }^{2}2x\left(2\right)dx=-\frac{\cot 2x}{2}+c$

$\int {\tan }^{2}8xdx$

$\int {\tan }^{2}8xdx=\int ({\sec }^{2}8x-1)dx=\frac{\tan 8x}{8}-x+c$

$\int \frac{\sqrt{\cot 2x}}{1-{\cos }^{2}2x}dx$

$$\begin{gathered} \int \frac{\sqrt{\cot 2x}}{1-{\cos }^{2}2x}dx=\int \frac{\sqrt{\cot 2x}}{{\sin }^{2}2x} \\ \left(f\left(x\right)=\cot 2x\Rightarrow f\text{'}\left(x\right)=-2{\csc }^{2}2x\right) \\ \begin{array}{rl}& =-\frac{1}{2}\int (\cot 2x{)}^{\frac{1}{2}}({\csc }^{2}2x)(-2)dx\\ & =-\frac{1}{2}\frac{(\cot 2x{)}^{\frac{3}{2}}}{\frac{3}{2}}+c=\frac{-1}{3}(\cot 2x{)}^{\frac{3}{2}}+c=\frac{-1}{3}\sqrt{({\cot }^{3}2x)}+c\end{array} \end{gathered}$$

$\int {\cos }^{2}2xdx$

$\begin{array}{rl}\int {\cos }^{2}2xdx& =\int \frac{1}{2}(1+\cos 4x)dx=\frac{1}{2}[\int dx+\frac{1}{4}\int \cos 4x(4\left)dx\right]\\ & =\frac{1}{2}(x+\frac{\sin 4x}{4})+c=\frac{1}{2}x+\frac{\sin 4x}{8}+c\end{array}$

$\int {\sin }^{2}8xdx$

$\int {\sin }^{2}8xdx=\int \frac{1}{2}(1-\cos 16x)dx=\frac{1}{2}(x-\frac{\sin 16x}{16})+c=\frac{1}{2}x-\frac{\sin 16x}{32}$

$\int {\cos }^{4}3xdx$

$$\begin{gathered} \int {\cos }^{4}3xdx=\int {({\cos }^{2}3x)}^{2}dx=\int {\left[\frac{1}{2}\right(1+\cos 6x\left)\right]}^{2}dx \\ \begin{array}{rl}& =\frac{1}{4}\int (1+2\cos 6x+{\cos }^{2}6x)dx\left({\cos }^{2}6x=\frac{1}{2}(1+\cos 12x)\right)\\ & =\frac{1}{4}\int (1+2\cos 6x+\frac{1}{2}(1+\cos 12x\left)\right)dx\\ & =\frac{1}{4}\int dx+\frac{1}{6}\int 2\cos 6x\left(6\right)dx+\frac{1}{2}[\int dx+\frac{1}{12}\int \cos 12x(12\left)dx\right]\end{array} \\ \begin{array}{rl}& =\frac{1}{4}[x+\frac{2\sin 6x}{6}+\frac{1}{2}(x+\frac{\sin 12x}{12})]+c\\ & =\frac{1}{4}x+\frac{1}{12}\sin 6x+\frac{1}{8}x+\frac{1}{96}\sin 12x+c\\ & =\frac{3}{8}x+\frac{1}{12}\sin 6x+\frac{1}{96}\sin 12x+c\end{array} \end{gathered}$$