L'intégrale simple

\(\boxed{I_1 = \int \sin^3x \cos^2x dx}\)

\(I_1 = \int \sin^2x \cos^2x \sin x dx = \int (1 - \cos^2x) \cos^2x \sin x dx\)

Posons \(u = \cos x \Leftrightarrow du = - \sin x dx\)

d'où \(I_1=-\int(1-u^2)u^2du=-\int(u^2-u^4)du=-\frac{u^3}{3}+\frac{u^5}{5}+C\)

\(\color{red}I_1=-\frac13\cos^3x+\frac15\cos^5x+C\)

\(\boxed{I_2 = \int \sin^2x \cos^3x dx}\)

\(I_2 = \int \sin^2x \cos^2x \cos x dx = \int \sin ^2x (1 -\sin ^2x) \cos x dx\)

Posons \(u = \sin x \Leftrightarrow du = \cos x dx\)

d'où \(I_2=\int u^2(1-u^2)du=\int(u^2-u^4)du=\frac{u^3}{3}-\frac{u^5}{5}+C\)

\(\color{red}I_2=\frac13\sin^3x-\frac15\sin^5x+C\)

\(\boxed{I_3 = \int \sin^3x \cos x dx}\)

\(I_3=\int\sin^2x\sin x\cos xdx=\int(\frac{1-\cos2x}{2})\frac12\sin2xdx\)

Posons \(u = \cos 2x \Leftrightarrow du = -2 \sin 2x dx\)

d'où

\(I_3=-\frac18\int(1-u)du=-\frac18(u-\frac{u^2}2)+C\)

\(\color{red}I_3=-\frac18(\cos^22x-\frac12\cos^2x)+C\)

\(\boxed{I_4 = \int \sin^2x \cos^2x dx}\)

\(I_4=\int\sin^2x\cos^2xdx=\frac14\int\sin^22xdx=\frac14\int\frac{1-\cos4x}2dx\)

\(=\frac18[\int dx-\int\cos4xdx]\)

d'où : \(\color{red}I_4\color{black}=\frac18x-\frac{1}{32}\sin4x+C\)

\(\boxed{I_5 = \int \sin 2x \cos 3x dx}\)

\(\sin2x\cos3x=\frac12[\sin(2x+3x)+\sin(2x-3x)]=\frac12(\sin5x-\sin x)\)

d'où \(I_5=\frac12\int(\sin5x-\sin x)dx=\color{red}-\frac1{10}\cos5x+\frac12\cos x+C\)

\(\boxed{I_6 = \int \sin 3x \sin 2x dx}\)

\(\sin3x\sin2x=\frac12[\cos(3x-2x)-\cos(3x+2x)]=\frac12(\cos x-\cos5 x)\)

d'où \(I_6=\frac12\int(\cos x-\cos5 x)dx=\color{red}\frac12\sin x-\frac1{10}\sin5 x+C\)

\(\boxed{I_7 = \int \cos 3x \cos 4x dx}\)

\(\cos3x\cos4x=\frac12[\cos(3x+4x)+\cos(3x-4x)]\\=\frac12(\cos7x+\cos x)\)

d'où \(I_7=\frac12\int(\cos7x+\cos x)dx=\color{red}\frac12\sin x+\frac1{14}\sin7x+C\)

\(\boxed{I_8=\int\frac{\sin x}{1+\cos x}dx~~(x\neq(2k+1)\pi)~~k\in\mathbb Z}\)

Posons \(t = \tan(x / 2)\) d'où

\(x=2\arctan t\Leftrightarrow dx=\frac{2dt}{1+t^2}\)

avec \(\sin t = \frac{2t}{1+t^2}\) et \(\cos t=\frac{1-t^2}{1+t^2}\)

et

\(I_8=\int\frac{2t}{(1+t^2)(1+\frac{1-t^2}{1+t^2})}\frac{2}{1+t^2}dt\)

\(=\int\frac{2t}{1+t^2}dt=\ln(1+t^2)+C\)

\(\color{red}I_8=\ln(1+\tan^2\frac{x}{2})+C_1\) (avec \(C_1=C+\ln 2)\)

Autres expressions :

\(1+\tan^2 \frac x2=\frac1{\cos^2 \frac x2}\Rightarrow I_8=-\ln\cos^2\frac x2+C\)

ou \(I_8=-\ln\frac{|1+\cos x|}2+C\) ou \(I_8=-\ln|1+\cos x|+C\)

\(\boxed{I_8=\int\frac{\sin x}{1+\cos x}dx~~(x\neq(2k+1)\pi)~~k\in\mathbb Z}\)

Posons \(\omega(x)=\frac{\sin x}{1+\cos x}dx\)l'élément différentiel.

Comme\(\omega(-x)=\frac{\sin(-x)d(-x)}{1+\cos(-x)}=\frac{\sin xdx}{1+\cos x}=\omega(x)\)

Posons \(\color{blue}t = \cos x\) d'où \(\color{blue}dt = - \sin x dx\) alors :

\(I_8=\int-\frac{dt}{1+t}=-\ln|1+t|+C\)

\(\color{red}I_8=-\ln|1+\cos x|+C\)

\(\boxed{I_9=\int\frac{\sin x\cos x}{\sin x + 1}dx~~(x\neq\frac{3\pi}{2}+k\pi)~~k\in \mathbb Z}\)

Posons \(\omega(x)=\frac{\sin x\cos x}{\sin x + 1}dx\)l'élément différentiel

Comme

\(\omega(\pi-x)=\frac{\sin (\pi-x)\cos (\pi-x)d(\pi-x)}{\sin(\pi- x) + 1}\)

\(=\frac{\sin x(-\cos x)d(-x)}{\sin x+1}=\omega(x)\)

Posons \(\color{blue}t = \sin x\) d'où \(\color{blue}dt = \cos x dx\) alors :

\(I_9=\int \frac t{t+1}dt=\int\frac{t+1-1}{t+1}dt\\=\int dt-\int\frac{dt}{t+1}\\=t-\ln|t+1|+C\)

d'où \(\color{red}I_9 = \sin x - \ln (1 + \sin x) + C\)

\(\boxed{I_{10}=\int\frac{dx}{1+\sin^2x}}\)

Posons \(\omega(x)=\frac{dx}{1+\sin^2x}\)l'élément différentiel

Comme

\(\omega(\pi+x)=\frac{d(\pi+x)}{1+\sin^2(\pi+x)}\\=\frac{dx}{1+(-\sin x)^2}=\omega(x)\)

Posons \(\color{blue}t = \tan x\) d'où \(\color{blue}dt = (1 + \tan^2 x) dx = (1 + t^2) dx\) alors :

\(I_{10}=\int\frac{dt}{(1+t^2)[1+\frac{t^2}{1+t^2}]}=\int\frac{dt}{1+2t^2}\)

\(=\frac1{\sqrt2}\arctan t\sqrt2+C\)

\(\color{red}I_{10}=\frac{1}{\sqrt 2}\arctan(\sqrt2\tan x)+C\)

\(\boxed{I=\int\frac{dx}{\cos^2x}}\)

\(n = 2\) est pair :

Posons \(\color{blue}t = \tan x\color{black}\Leftrightarrow \color{blue}dt\color{black} = (1 + \tan^2 x) dx =\color{blue} dx / \cos^2 x\)

d'où \(\color{red}I\color{black} = \int dt = t + C =\color{red} \tan x + C\)

\(\boxed{J=\int\frac{dx}{\sin x}}\)

\(n = 1\) est impair :

Posons \(\color{blue}t = \tan (x / 2)\color{black} ⇔ x = 2 \arctan t\) et \(\color{blue}dx = 2dt / (1 + t^2)\)

Sachant que \(\sin x = 2t / (1 + t^2),\) nous obtenons :

\(\color{red}J\color{black}=\int\frac{2dt}{(1+t^2)\frac{2t}{1+t^2}}=\int\frac{dt}{t}\\=\ln|t|+C=\color{red}\ln|\tan\frac{x}{2}|+C\)

\(\boxed{K = \int \tan^2 x dx}\)

\(n = 2\) est pair et positif:

Ajoutons et retranchons \(1\) :

\(K = \int (\tan^2 x + 1 - 1) dx\\= \int (\tan^2 x + 1) dx - \int dx\)

\(\color{red}K = \tan x - x + C\)

\(\boxed{L=\int\frac{1}{\tan x}dx}\)

\(n = -1\) est impair et négatif:

Posons \(\color{blue}t = \sin x \color{black}\Leftrightarrow \color{blue}dt = \cos x dx\)

\(\color{red}L\color{black}=\int\frac1{\tan x}dx=\int\frac{\cos x}{\sin x}dx\\=\frac{dt}{t}=\color{red}\ln|t|+C\)

\(\color{red}L = \ln |\tan x| + C\)

\(\boxed{I_{11}=\int_0^1\frac{dx}{(1+x^2)^2}}\)

Posons \(\color{blue}x = \tan t\) d'où \(\color{blue}t = \arctan x\)

et \(\color{blue}dt = dx / (1 + x^2)\) avec les changements de bornes :

\(t_1 = \arctan 0 = 0\)

\(t_2 = \arctan 1 = \pi / 4\)

d'où

\(\color{red}I_{11}\color{black}=\int_0^{\pi/4}\frac{(1+\tan^2t)dt}{(1+\tan^2t)^2}=\int_0^{\pi/4}\frac{dt}{(1+\tan^2t)}\\\int_0^{\pi/4}\cos^2tdt=\int_0^{\pi/4}\frac{1+\cos2t}{2}dt\\=[\frac t2+\frac{\sin 2t}{4}]_0^{\pi/4}=\frac{\pi}8+\frac14=\color{red}\frac{\pi+2}{8}\)