جد تكاملات كلاً مما يأتي:

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

$$\begin{gathered} \begin{array}{rl}& =\int (sin2xco{s}^{2}2x+2sin2x-co{s}^{2}2x-2)dx\\ & =\int \left[\right(cos2x{)}^{2}sin2x+2sin2x-co{s}^{2}2x-2]dx\\ & \left(-2\sin 2x=\left(\mathrm{المشتقة}\right) {\cos }^{2}2x=\frac{1}{2}(1+\cos 4x)\right)\end{array} \\ \begin{array}{rl}& =\int [-\frac{1}{2}(\cos 2x{)}^2\cdot (-2\sin 2x)+\sin 2x\left(2\right)-\frac{1}{2}(1+\cos 4x)-2]dx\\ & =-\frac{1}{2}\frac{(\cos 2x{)}^3}{3}-\cos 2x-\frac{1}{2}(x+\frac{\sin 4x}{4})-2x+c\\ & =-\frac{(\cos 2x{)}^3}{6}-\cos 2x-\frac{1}{2}x+\frac{\sin 4x}{8}-2x+c\\ & =-\frac{(\cos 2x{)}^3}{6}-\cos 2x-\frac{5}{2}x+\frac{\sin 4x}{8}+c\end{array} \end{gathered}$$

التمارين العامة الخاصة بالفصل الرابع

التمارين العامة الخاصة بالفصل الرابع

(1)- جد تكاملات كلاً مما يأتي:

$\int ({\cos }^{4}x-{\sin }^{4}x)dx$

نحلل فرق بين مربعين

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

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

$$\begin{gathered} \begin{array}{rl}& =\int (sin2xco{s}^{2}2x+2sin2x-co{s}^{2}2x-2)dx\\ & =\int \left[\right(cos2x{)}^{2}sin2x+2sin2x-co{s}^{2}2x-2]dx\\ & \left(-2\sin 2x=\left(\mathrm{المشتقة}\right) {\cos }^{2}2x=\frac{1}{2}(1+\cos 4x)\right)\end{array} \\ \begin{array}{rl}& =\int [-\frac{1}{2}(\cos 2x{)}^2\cdot (-2\sin 2x)+\sin 2x\left(2\right)-\frac{1}{2}(1+\cos 4x)-2]dx\\ & =-\frac{1}{2}\frac{(\cos 2x{)}^3}{3}-\cos 2x-\frac{1}{2}(x+\frac{\sin 4x}{4})-2x+c\\ & =-\frac{(\cos 2x{)}^3}{6}-\cos 2x-\frac{1}{2}x+\frac{\sin 4x}{8}-2x+c\\ & =-\frac{(\cos 2x{)}^3}{6}-\cos 2x-\frac{5}{2}x+\frac{\sin 4x}{8}+c\end{array} \end{gathered}$$

$\int \frac{\ln \left(x\right)}{x}dx$

$\int \frac{\ln \left(x\right)}{x}dx=\int \ln x.\underset{\left(\ln \mathrm{مشتقة}\right)}{\underbrace{\frac{1}{x}}}dx=\frac{(\ln x{)}^2}{2}+c$

$\int \frac{2\sin \sqrt[3]{x}}{\sqrt[3]{x^2}}dx$

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

$\int \cot x{\csc }^{3}xdx$

$\begin{array}{rl}& =\int (\csc x{)}^3\cot xdx\left(\csc \left(\mathrm{مشتقة}\right)=-\csc x\cdot \cot x\right)\\ & =-\int (\csc x{)}^2(-\csc x\cdot \cot x)dx=-\frac{(\csc x{)}^3}{3}+c\end{array}$

$\int \sqrt[3]{3x^3-5x^5}dx$

$\begin{array}{rl}& \int \sqrt[3]{3x^3-5x^5}dx\\ & =\int \sqrt[3]{x^3(3-5x^2)}dx\\ & =\int x\sqrt[3]{(3-5x^2)}dx\\ & =\int {(3-5x^2)}^{\frac{1}{3}}xdx=-\frac{1}{10}\int {(3-5x^2)}^{\frac{1}{3}}(-10x)dx\\ & =-\frac{1}{10}\frac{{(3-5x^2)}^{\frac{4}{3}}}{\frac{4}{3}}+c=-\frac{3}{40}{(3-5x^2)}^{\frac{4}{3}}+c\end{array}$

$\int \frac{1}{x^2-14x+49}dx$

نحلل المقام مربع كامل

$\int \frac{1}{(x-7{)}^2}dx=\int (x-7{)}^{-2}=\frac{(x-7{)}^{-1}}{-1}+c=\frac{1}{-(x-7)}+c$

$\int {\sec }^{2}3x\cdot e^{\tan 3x}dx$

$$\begin{gathered} \int {\sec }^{2}3x\cdot e^{\tan 3x}dx=\int e^{\tan 3x}\cdot {\sec }^{2}3xdx=\frac{1}{3}\int e^{\tan 3x}\cdot (3{\sec }^{2}3x)dx \\ =\frac{1}{3}{e}^{\tan 3x}+c \end{gathered}$$