تكامل الدوال المثلثية التربيعية

1- التكاملات المباشرة كما في الأمثلة التالية:

$\int {\sec }^2\theta d\theta$

$\int {\sec }^2\theta d\theta =\tan \theta +c$

$\int {\csc }^2\theta d\theta$

$\int {\csc }^2\theta d\theta =-\cot \theta +c$

$\int {\tan }^2\theta d\theta$

$\int {\tan }^2\theta d\theta =\int ({\sec }^2\theta -1)d\theta =\int {\sec }^2\theta d\theta -\int d\theta =\tan \theta -\theta +c$

$\int co{t}^2\theta d\theta$

$\int {\cot }^2\theta d\theta =\int ({\csc }^2\theta -1)d\theta =\int {\csc }^2\theta d\theta -\int d\theta =-\cot \theta -\theta +c$

$\int {\sin }^2\theta d\theta$

$$\begin{gathered} \int {\sin }^2\theta d\theta =\int \frac{1-\cos 2\theta }{2}d\theta =\frac{1}{2}[\int d\theta -\frac{1}{2}\int \cos 2\theta \underset{\mathrm{الزاوية} \mathrm{مشتقة}}{\underbrace{\left(2\right)}}d\theta ] \\ =\frac{1}{2}(\theta -\frac{1}{2}\sin 2\theta )+c=\frac{1}{2}\theta -\frac{1}{4}\sin 2\theta +c \end{gathered}$$

$\int {\cos }^2\theta d\theta$

$$\begin{gathered} \int {\cos }^2\theta d\theta =\int \frac{1+\cos 2\theta }{2}d\theta =\frac{1}{2}[\int d\theta +\frac{1}{2}\int \cos 2\theta \underset{\text{الزاوية مشتقة}}{\underbrace{\left(2\right)}}d\theta ] \\ =\frac{1}{2}(\theta +\frac{1}{2}\sin 2\theta )+c=\frac{1}{2}\theta +\frac{1}{4}\sin 2\theta +c \end{gathered}$$

$\int \sin (ax+b)$

$\int \sin (ax+b)=-\frac{1}{a}\cos (ax+b)$

$\int \cos (ax+b)$

$\int \cos (ax+b)=\frac{1}{a}\sin (ax+b)$

2- التكاملات باستخدام قاعدة مشتقة الزاوية للدالة المثلثية وكما في الأمثلة التالية:

(1)- جد التكاملات لكل مما يأتي:

$\int 9\sin 3xdx$

$\int 9\sin 3xdx=3\int \underset{\mathrm{الزاوية} \mathrm{مشتقة}}{\underbrace{\left(3\right)}} \sin 3xdx=-3\cos 3x+c$

$\int x^2\sin x^{3}dx$

$\int x^2\sin x^{3}dx=\frac{1}{3}\int \underset{\mathrm{الزاوية} \mathrm{مشتقة}}{\underbrace{\left(3x^2\right)}}\sin x^{3}dx=-\frac{1}{3}\cos x^3+c$

$\int \sqrt{1-\sin 2x}dx$

$\begin{array}{rl}& \int \sqrt{1-\sin 2x}dx=\int \sqrt{{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x}dx\\ & =\int \sqrt{{\sin }^{2}x-2\sin x\cos x+{\cos }^{2}x}dx=\int \sqrt{(\sin x-\cos x{)}^2}dx\\ & =\int \pm (\sin x-\cos x)dx\\ & =\pm (-\cos x-\sin x)+c=\mp (\cos x+\sin x)+c\end{array}$

ملاحظة:

$\int {\sec }^2(ax+b)=\frac{1}{a}\tan (ax+b)$

$\int (\sin x-\cos x{)}^7(\cos x+\sin x)dx$

$\int (\sin x-\cos x{)}^7(\cos x+\sin x)dx=\frac{(\sin x-\cos x{)}^8}{8}+c$

$\int \frac{1+{\tan }^{2}x}{{\tan }^{3}x}dx$

$\int \frac{1+{\tan }^{2}x}{{\tan }^{3}x}dx=\int \frac{{\sec }^{2}x}{{\tan }^{3}x}dx=\int {\tan }^{-3}{\sec }^{2}xdx=\frac{{\tan }^{-2}x}{-2}+c=\frac{-1}{2{\tan }^{2}x}+c$

ملاحظة:

$\int \sec \mathrm{ax}\tan ax=\frac{1}{a}\sec ax\int \csc ax\cot ax=-\frac{1}{a}\mathrm{cscax}$

$\int \frac{\tan x}{{\cos }^{2}x}dx$

$\int \frac{\tan x}{{\cos }^{2}x}dx=\int \tan x{\sec }^{2}xdx=\frac{{\tan }^{2}x}{2}+c$

$\begin{array}{rl}& \int \sin 6x{\cos }^{2}3xdx\end{array}$

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

$\int \frac{\cos 4x}{\cos 2x-\sin 2x}dx$

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

$\begin{array}{rl}& \int {\cot }^{2}5xdx\end{array}$

$\begin{array}{rl}& \int {\cot }^{2}5xdx=\int ({\csc }^{2}5x-1)dx=\int {\csc }^{2}5x-\int dx=\frac{1}{5}\int {\csc }^{2}5x\left(5\right)dx-\int dx\\ & =-\frac{1}{5}\cot 5x-x+c\end{array}$

$\int ta{n}^{2}7xdx$

$\int {\tan }^{2}7xdx=\int ({\sec }^{2}7x-1)dx=\frac{1}{7}\int ({\sec }^{2}7x)\left(7\right)dx-\int dx=\frac{1}{7}\tan 7x-x+c$

3- إذا كانت الزوايا في السؤال غير متساوية فنساوي الزوايا كما يلي:

$\int \sin 4x\cdot {\cos }^{2}2xdx$

$\begin{array}{rl}& =\int 2\sin 2x\cos 2x{\cos }^{2}2xdx=2\int (\cos 2x{)}^3\sin 2xdx\\ & =\frac{2}{-2}\int (\cos 2x{)}^3(\sin 2x\left)\right(-2)dx=-\frac{(\cos 2x{)}^4}{4}+c\end{array}$

ملاحظة:

$$\begin{gathered} \sin 4x=2\sin 2x\cos 2x \\ \sin 2x=2\sin x\cos x \end{gathered}$$

$\int \sin 6x\cdot {\cos }^{4}3xdx$

$\begin{array}{rl}& \int 2\sin 3x\cos 3x\cdot {\cos }^{4}3xdx=2\int {\cos }^{5}3x\sin 3xdx\\ & =\frac{-2}{3}\int {\cos }^{5}3x(-3\sin 3x)dx=\frac{-2}{3}\frac{{\cos }^{6}3x}{6}+c=\frac{-}{9}{\cos }^{6}3x+c\end{array}$

3- تكاملات مربعات الدوال الدائرية:

${\cos }^{2}x=\{\begin{array}{ll}1-{\sin }^{2}x& \text{فردي }n\\ \frac{1}{2}(1+\cos 2x)& \text{زوجي }n\end{array}$

${\sin }^{2}x=\{\begin{array}{ll}1-{\cos }^{2}x& \mathrm{فردي} n\\ \frac{1}{2}(1-\cos 2x)& \text{زوجي }n\end{array}$

(2)- جد تكامل كلاً مما يأتي:

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

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

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

$$\begin{gathered} \int {({\cos }^{2}x)}^{2}dx=\int {\left[\frac{1}{2}\right(1+\cos 2x\left)\right]}^{2}dx=\frac{1}{4}\int (1+\cos 2x{)}^{2}dx\left({\cos }^{2}x=\frac{1}{2}(1+\cos 2x)\right) \\ \begin{array}{rl}& =\frac{1}{4}\int (1+2\cos 2x+{\cos }^{2}2x)=\frac{1}{4}\int [1+2\cos 2x+\frac{1}{2}(1+\cos 4x\left)\right]dx\\ & =\frac{1}{4}[x+\sin 2x+\frac{1}{2}(x+\frac{\sin 4x}{4})]+c=\frac{1}{4}x+\frac{1}{4}\sin 2x+\frac{1}{8}x+\frac{1}{32}\sin 4x+c\end{array} \end{gathered}$$

$\begin{array}{rl}& \int {\sin }^{2}3xdx\end{array}$

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

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

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

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

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

$\int {\sin }^{4}xdx$

$$\begin{gathered} \int {\sin }^{4}xdx=\int {({\sin }^{2}x)}^{2}dx=\int {\left[\frac{1-\cos 2x}{2}\right]}^{2}dx=\int \frac{(1-2\cos 2x+{\cos }^{2}2x)}{4}dx \\ =\int \frac{1}{4}(1-2\cos 2x+{\cos }^{2}2x)dx \\ \begin{array}{rl}& =\frac{1}{4}(\int dx-\int 2\cos 2xdx+\int {\cos }^{2}2xdx)\\ & =\frac{1}{4}(\int dx-\int 2\cos 2xdx+\int \frac{1}{2}(1+\cos 4x\left)dx\right)\\ & =\frac{1}{4}(\int dx-\int 2\cos 2xdx+\frac{1}{2}[\int dx+\frac{1}{4}\int \cos 4x\cdot (4\left)\right]dx)\\ & =\frac{1}{4}(x-\sin 2x+\frac{1}{2}x+\frac{1}{8}\sin 4x)+c=\frac{1}{4}x-\frac{1}{4}\sin 2x+\frac{1}{8}x+\frac{1}{32}\sin 4x+c\\ & =\frac{3}{8}x-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+c\end{array} \end{gathered}$$

$\int \cos 2x\cdot {\sin }^{2}xdx$

$$\begin{gathered} \left({\cos }^{2}2x=\frac{1}{2}(1+\cos 4x){\sin }^{2}x=\frac{1}{2}(1-\cos 2x)\right) \\ \begin{array}{rl}& =\int \cos 2x\frac{1}{2}(1-\cos 2x)dx=\frac{1}{2}\int (\cos 2x-{\cos }^{2}2x)dx\\ & =\frac{1}{2}\int [\cos 2x-\frac{1}{2}(1+\cos 4x\left)\right]dx=\frac{1}{2}[\frac{1}{2}\int (\cos 2x(2)-\frac{1}{2}[\int dx+\frac{1}{4}\int \cos 4x(4\left)dx\right]]\\ & =\frac{1}{2}[\frac{\sin 2x}{2}-\frac{1}{2}(x+\frac{\sin 4x}{4})]+c=\frac{1}{4}\sin 2x-\frac{1}{4}x+\frac{\sin 4x}{16}+c\end{array} \end{gathered}$$

$\int {\cos }^{3}2xdx$

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

$\int \sqrt{1+\sin 2x}dx$

$\begin{array}{rl}& \int \sqrt{1+\sin 2x}dx=\int \sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}dx\\ & =\int \sqrt{{\sin }^{2}x+2\sin x\cos x+{\cos }^{2}x}dx=\int \sqrt{(\sin x+\cos x{)}^2}dx\\ & =\int \pm (\sin x+\cos x)dx=\pm (-\cos x+\sin x)+c=\mp (\cos x-\sin x)+c\end{array}$

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

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

$\int (\cos 2x-\sec x)(\cos 2x+\sec x)dx$

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

$\int \sqrt[3]{x^3+2x^5}dx$

نستخرج عامل مشترك لأن المشتقة داخل القوس غير موجودة.

$$\begin{gathered} =\int \sqrt[3]{x^3(1+2x^2)}dx=\int x\sqrt[3]{1+2x^2}dx=\int x{(1+2x^2)}^{\frac{1}{3}}dx \\ =\frac{1}{4}\int {(1+2x^2)}^{\frac{1}{3}}\cdot 4xdx=\frac{1}{4}\frac{{(1+2x^2)}^{\frac{4}{3}}}{\frac{4}{2}}+c=\frac{3}{16}{(1+2x^2)}^{\frac{4}{3}}+c \end{gathered}$$

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

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

$\int (\tan x+\cot x{)}^{2}dx$

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

ملاحظة:

$\{2{\sin }^{2}x-1 , 1-2{\cos }^{2}x , {\sin }^{2}x-{\cos }^{2}x , {\sin }^{4}x-{\cos }^4\}=-\cos 2x$