حلول أسئلة الصف السادس الإعدادي

حل اسئلة رياضيات - علوم - عربي وجميع الكتب والمواد الأخرى

تكامل الدوال المثلثية التربيعية

1- التكاملات المباشرة كما في الأمثلة التالية:

$\int \sec^{2} θ d θ$

$\int \sec^{2} θ d θ = \tan θ + c$

$\int \csc^{2} θ d θ$

$\int \csc^{2} θ d θ = - \cot θ + c$

$\int \tan^{2} θ d θ$

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

$\int c o t^{2} θ d θ$

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

$\int \sin^{2} θ d θ$

$\int \sin^{2} θ d θ = \int \frac{1 - \cos 2 θ}{2} d θ = \frac{1}{2} [\int d θ - \frac{1}{2} \int \cos 2 θ \underset{الزاوية مشتقة}{\underset{⏟}{(2)}} d θ] = \frac{1}{2} (θ - \frac{1}{2} \sin 2 θ) + c = \frac{1}{2} θ - \frac{1}{4} \sin 2 θ + c$

$\int \cos^{2} θ d θ$

$\int \cos^{2} θ d θ = \int \frac{1 + \cos 2 θ}{2} d θ = \frac{1}{2} [\int d θ + \frac{1}{2} \int \cos 2 θ \underset{الزاوية مشتقة}{\underset{⏟}{(2)}} d θ] = \frac{1}{2} (θ + \frac{1}{2} \sin 2 θ) + c = \frac{1}{2} θ + \frac{1}{4} \sin 2 θ + c$

$\int \sin (a x + b)$

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

$\int \cos (a x + b)$

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

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

(1)- جد التكاملات لكل مما يأتي:

$\int 9 \sin 3 x d x$

$\int 9 \sin 3 x d x = 3 \int \underset{الزاوية مشتقة}{\underset{⏟}{(3)}} \sin 3 x d x = - 3 \cos 3 x + c$

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

$\int x^{2} \sin x^{3} d x = \frac{1}{3} \int \underset{الزاوية مشتقة}{\underset{⏟}{(3 x^{2})}} \sin x^{3} d x = - \frac{1}{3} \cos x^{3} + c$

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

$\begin{aligned} \int \sqrt{1 - \sin 2 x} d x = \int \sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cos x} d x \\ = \int \sqrt{\sin^{2} x - 2 \sin x \cos x + \cos^{2} x} d x = \int \sqrt{(\sin x - \cos x)^{2}} d x \\ = \int \pm (\sin x - \cos x) d x \\ = \pm (- \cos x - \sin x) + c = \mp (\cos x + \sin x) + c \end{aligned}$

ملاحظة:

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

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

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

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

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

ملاحظة:

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

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

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

$\begin{aligned} \int \sin 6 x \cos^{2} 3 x d x \end{aligned}$

$\begin{aligned} \int \sin 6 x \cos^{2} 3 x d x = \int (2 \sin 3 x \cos 3 x) \cos^{2} 3 x d x \\ = \frac{2}{- 3} \int \cos^{3} 3 x (\sin 3 x) (- 3) d x = (\frac{- 2}{3}) \frac{\cos^{4} 3 x}{4} + c = \frac{- 1}{6} \cos^{4} 3 x + c \end{aligned}$

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

$\begin{aligned} \int \frac{\cos 4 x}{\cos 2 x - \sin 2 x} d x = \int \frac{\cos^{2} 2 x - \sin^{2} 2 x}{\cos 2 x - \sin 2 x} d x \\ = \int \frac{(\cos 2 x + \sin 2 x) (\cos 2 x - \sin 2 x)}{\cos 2 x - \sin 2 x} d x \\ = \int (\cos 2 x + \sin 2 x) d x = \frac{1}{2} \int \cos 2 x (2) d x + \frac{1}{2} \int (\sin 2 x) (2) d x \\ = \frac{1}{2} \sin 2 x - \frac{1}{2} \cos 2 x + c \end{aligned}$

$\begin{aligned} \int \cot^{2} 5 x d x \end{aligned}$

$\begin{aligned} \int \cot^{2} 5 x d x = \int (\csc^{2} 5 x - 1) d x = \int \csc^{2} 5 x - \int d x = \frac{1}{5} \int \csc^{2} 5 x (5) d x - \int d x \\ = - \frac{1}{5} \cot 5 x - x + c \end{aligned}$

$\int t a n^{2} 7 x d x$

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

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

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

$\begin{aligned} = \int 2 \sin 2 x \cos 2 x \cos^{2} 2 x d x = 2 \int (\cos 2 x)^{3} \sin 2 x d x \\ = \frac{2}{- 2} \int (\cos 2 x)^{3} (\sin 2 x) (- 2) d x = - \frac{(\cos 2 x)^{4}}{4} + c \end{aligned}$

ملاحظة:

$\sin 4 x = 2 \sin 2 x \cos 2 x \sin 2 x = 2 \sin x \cos x$

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

$\begin{aligned} \int 2 \sin 3 x \cos 3 x \cdot \cos^{4} 3 x d x = 2 \int \cos^{5} 3 x \sin 3 x d x \\ = \frac{- 2}{3} \int \cos^{5} 3 x (- 3 \sin 3 x) d x = \frac{- 2}{3} \frac{\cos^{6} 3 x}{6} + c = \frac{-}{9} \cos^{6} 3 x + c \end{aligned}$

3- تكاملات مربعات الدوال الدائرية:

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

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

(2)- جد تكامل كلاً مما يأتي:

$\int \cos^{3} x d x$

$\begin{aligned} \int \cos^{3} x d x & = \int \cos^{2} x \cos x d x = \int (1 - \sin^{2} x) \cdot \cos x d x (\cos^{2} x = 1 - \sin^{2} x) \\ = \int \cos x - \sin^{2} x \cos x d x = \sin x - \frac{(\sin x)^{3}}{2} + c \end{aligned}$

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

$\int {(\cos^{2} x)}^{2} d x = \int {[\frac{1}{2} (1 + \cos 2 x)]}^{2} d x = \frac{1}{4} \int (1 + \cos 2 x)^{2} d x (\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)) \begin{aligned} = \frac{1}{4} \int (1 + 2 \cos 2 x + \cos^{2} 2 x) = \frac{1}{4} \int [1 + 2 \cos 2 x + \frac{1}{2} (1 + \cos 4 x)] d x \\ = \frac{1}{4} [x + \sin 2 x + \frac{1}{2} (x + \frac{\sin 4 x}{4})] + c = \frac{1}{4} x + \frac{1}{4} \sin 2 x + \frac{1}{8} x + \frac{1}{32} \sin 4 x + c \end{aligned}$

$\begin{aligned} \int \sin^{2} 3 x d x \end{aligned}$

$\begin{aligned} \int \sin^{2} 3 x d x = \int \frac{(1 - \cos 6 x)}{2} d x = \frac{1}{2} \int d x - \int \frac{\cos 6 x}{2} d x \\ = \frac{1}{2} \int d x - \frac{1}{6} \int \frac{1}{2} \cos 6 x (6) d x = \frac{1}{2} x - \frac{1}{12} \sin 6 x + c \end{aligned}$

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

$\begin{aligned} \int (\cos^{4} x - \sin^{4} x) d x = \int (\cos^{2} x - \sin^{2} x) (\cos^{2} x + \sin^{2} x) d x \\ (\cos^{2} x - \sin^{2} x = \cos 2 x, \cos^{2} x + \sin^{2} x = 1) \\ = \int \cos 2 x d x = \frac{1}{2} \int \cos 2 x (2) d x = \frac{\sin 2 x}{2} + c \end{aligned}$

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

$\begin{aligned} \int \frac{\sin^{3} x d x}{1 - \cos x} = \int \frac{\sin^{2} x \sin x d x}{1 - \cos x} = \int \frac{(1 - \cos^{2} x) \sin x d x}{1 - \cos x} \\ = \int \frac{(1 - \cos x) (1 + \cos x) \sin x d x}{1 - \cos x} = \int (\sin x + \sin x \cos x) d x \\ = - \cos x + \frac{(\sin x)^{2}}{2} + c \end{aligned}$

$\int \sin^{4} x d x$

$\int \sin^{4} x d x = \int {(\sin^{2} x)}^{2} d x = \int {[\frac{1 - \cos 2 x}{2}]}^{2} d x = \int \frac{(1 - 2 \cos 2 x + \cos^{2} 2 x)}{4} d x = \int \frac{1}{4} (1 - 2 \cos 2 x + \cos^{2} 2 x) d x \begin{aligned} = \frac{1}{4} (\int d x - \int 2 \cos 2 x d x + \int \cos^{2} 2 x d x) \\ = \frac{1}{4} (\int d x - \int 2 \cos 2 x d x + \int \frac{1}{2} (1 + \cos 4 x) d x) \\ = \frac{1}{4} (\int d x - \int 2 \cos 2 x d x + \frac{1}{2} [\int d x + \frac{1}{4} \int \cos 4 x \cdot (4)] d x) \\ = \frac{1}{4} (x - \sin 2 x + \frac{1}{2} x + \frac{1}{8} \sin 4 x) + c = \frac{1}{4} x - \frac{1}{4} \sin 2 x + \frac{1}{8} x + \frac{1}{32} \sin 4 x + c \\ = \frac{3}{8} x - \frac{1}{4} \sin 2 x + \frac{1}{32} \sin 4 x + c \end{aligned}$

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

$(\cos^{2} 2 x = \frac{1}{2} (1 + \cos 4 x) \sin^{2} x = \frac{1}{2} (1 - \cos 2 x)) \begin{aligned} = \int \cos 2 x \frac{1}{2} (1 - \cos 2 x) d x = \frac{1}{2} \int (\cos 2 x - \cos^{2} 2 x) d x \\ = \frac{1}{2} \int [\cos 2 x - \frac{1}{2} (1 + \cos 4 x)] d x = \frac{1}{2} [\frac{1}{2} \int (\cos 2 x (2) - \frac{1}{2} [\int d x + \frac{1}{4} \int \cos 4 x (4) d x]] \\ = \frac{1}{2} [\frac{\sin 2 x}{2} - \frac{1}{2} (x + \frac{\sin 4 x}{4})] + c = \frac{1}{4} \sin 2 x - \frac{1}{4} x + \frac{\sin 4 x}{16} + c \end{aligned}$

$\int \cos^{3} 2 x d x$

$\begin{aligned} \int \cos^{3} 2 x d x = \int \cos 2 x \cos^{2} 2 x d x = \int \cos 2 x (1 - \sin^{2} 2 x) d x \\ = \int (\cos 2 x - \sin^{2} 2 x \cos 2 x) d x = \frac{1}{2} \int \cos 2 x (2) d x - \frac{1}{2} \int \sin^{2} 2 x \cos 2 x (2) d x \\ = \frac{1}{2} \sin 2 x - \frac{1}{2} (\frac{1}{3}) \sin^{3} 2 x + c = \frac{1}{2} \sin 2 x - \frac{1}{6} \sin^{3} 2 x + c \end{aligned}$

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

$\begin{aligned} \int \sqrt{1 + \sin 2 x} d x = \int \sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x} d x \\ = \int \sqrt{\sin^{2} x + 2 \sin x \cos x + \cos^{2} x} d x = \int \sqrt{(\sin x + \cos x)^{2}} d x \\ = \int \pm (\sin x + \cos x) d x = \pm (- \cos x + \sin x) + c = \mp (\cos x - \sin x) + c \end{aligned}$

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

$(\sin 2 x = 2 \sin x \cos x \sin x \cos x = \frac{1}{2} \sin x) \begin{aligned} \int \sin^{2} x \cos^{2} x d x & = \int (\sin x \cos x)^{2} d x = \int {[\frac{1}{2} \sin 2 x]}^{2} d x = \frac{1}{4} \int \sin^{2} 2 x d x \\ = \frac{1}{4} \int \frac{1}{2} (1 - \cos 4 x) d x = \frac{1}{8} (x - \frac{1}{4} \sin 4 x) + c \end{aligned}$

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

$\begin{aligned} \int (\cos 2 x - \sec x) (\cos 2 x + \sec x) d x = \int (\cos^{2} 2 x - \sec^{2} x) d x \\ = \int [\frac{1}{2} (1 + \cos 4 x) - \sec^{2} x] d x = \int \frac{1}{2} d x + \frac{1}{2} (\frac{1}{4}) \int \cos 2 x (4) d x - \int \sec^{2} x d x \\ = \frac{1}{2} x + \frac{1}{8} \sin 4 x - \tan x + c \end{aligned}$

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

نستخرج عامل مشترك لأن المشتقة داخل القوس غير موجودة.

$= \int \sqrt[3]{x^{3} (1 + 2 x^{2})} d x = \int x \sqrt[3]{1 + 2 x^{2}} d x = \int x {(1 + 2 x^{2})}^{\frac{1}{3}} d x = \frac{1}{4} \int {(1 + 2 x^{2})}^{\frac{1}{3}} \cdot 4 x d x = \frac{1}{4} \frac{{(1 + 2 x^{2})}^{\frac{4}{3}}}{\frac{4}{2}} + c = \frac{3}{16} {(1 + 2 x^{2})}^{\frac{4}{3}} + c$

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

$(\sin 2 x = 2 \sin x \cos x \sin x \cos x = \frac{1}{2} \sin x) \begin{aligned} \int (1 - 2 \sin 3 x)^{2} d x = \int (1 - 4 \sin 3 x + 4 \sin^{2} 3 x) d x \\ = \int (1 - 4 \sin 3 x + 4 (\frac{1}{2}) (1 - \cos 6 x)) d x = \int (1 - 4 \sin 3 x + 2 - 2 \cos 6 x) d x \\ = \int 3 d x - \frac{4}{3} \int \sin 3 x (3) d x - \frac{2}{6} \int \cos 6 x (6) d x = 3 x + \frac{4}{3} \cos 3 x - \frac{1}{3} \sin 6 x + c \end{aligned}$

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

$\begin{aligned} \int (\tan x + \cot x)^{2} d x = \int (\tan^{2} x + 2 \tan x \cot x + \cot^{2} x) d x \\ = \int (\sec^{2} x - 1 + 2 \frac{\sin x}{\cos x} \cdot \frac{\cos x}{\sin x} + (\csc^{2} x - 1)) d x \\ = \int (\sec^{2} x + \csc^{2} x - 2 + 2 (1)) d x \\ = \int (\sec^{2} x + \csc^{2} x) d x = \tan x - \cot x + c \end{aligned}$

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

حلول أسئلة الصف السادس الإعدادي

حل اسئلة رياضيات - علوم - عربي وجميع الكتب والمواد الأخرى

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حلول أسئلة الصف السادس الإعدادي

تكامل الدوال المثلثية التربيعية