Techniques d'intégration

\(u=x,\ v'=e^x\) : \([xe^x]_0^1-\int_0^1e^x=e-[e^x]_0^1=e-(e-1)\).

\(x^2+2x+5=(x+1)^2+4\).

\(u:1\to2\) : \(\int_1^2\tfrac{du}u=\ln2\).

\(I_1=\int_0^1e^{3t}dt\)

\(\tfrac{e^3-1}3\)

\(I_2=\int_1^{\sqrt3}\tfrac1{t^2+1}dt\)

\(\arctan\sqrt3-\arctan1\).

\(\tfrac\pi{12}\)

\(I_3=\int_0^{\pi/3}\cos(2u+\tfrac\pi3)du\)

\(\tfrac12[\sin\pi-\sin\tfrac\pi3]\).

\(-\tfrac{\sqrt3}4\)

\(I_4=\int_0^1 2xe^{x^2}dx\)

\(e-1\)

\(I_5=\int_0^1 xe^{x^2}dx\)

\(\tfrac{e-1}2\)

\(I_6=\int_0^1 x^3e^{x^2}dx\)

\(u=x^2\) : \(\tfrac12\int_0^1 ue^u=\tfrac12[(u-1)e^u]_0^1\).

\(\tfrac12\)

\(I_7=\int_1^2\tfrac{-2}{-2x+3}dx\)

\([\ln|-2x+3|]_1^2=\ln1-\ln1\).

\(0\)

\(I_8=\int_1^2\tfrac1{2-3x}dx\)

\(-\tfrac13[\ln|2-3x|]_1^2=-\tfrac13\ln4\).

\(-\tfrac23\ln2\)

\(f_1=\sin^5x\)

\(=\sin x(1-\cos^2x)^2\).

\(-\cos x+\tfrac23\cos^3x-\tfrac15\cos^5x\)

\(f_2=\cos^4t\)

\(=\tfrac{3+4\cos2t+\cos4t}8\).

\(\tfrac38t+\tfrac14\sin2t+\tfrac1{32}\sin4t\)

\(f_3=\cos^4x\sin^2x\)

Linéarisation.

\(\tfrac x{16}+\tfrac1{64}\sin2x-\tfrac1{64}\sin4x-\tfrac1{192}\sin6x\)

\(f_4=\cos(3x)\cos^3x\)

\(\cos^3x=\tfrac{3\cos x+\cos3x}4\).

\(\tfrac x8+\tfrac3{16}\sin2x+\tfrac3{32}\sin4x+\tfrac1{48}\sin6x\)

\(A=\int_{\pi/3}^{\pi/2}\cos x\sin^2x\,dx\) (\(u=\sin x\))

\([\tfrac{\sin^3x}3]\).

\(\tfrac13-\tfrac{\sqrt3}8\)

\(B=\int_1^4\tfrac{t^2}{\sqrt{t+1}}dt\) (\(u=\sqrt{t+1}\))

\(2\int_{\sqrt2}^{\sqrt5}(u^2-1)^2du\).

\(\tfrac{16\sqrt5}3-\tfrac{14\sqrt2}{15}\)

\(C=\int_1^4\tfrac{1-\sqrt t}{\sqrt t}dt\) (\(x=\sqrt t\))

\(2\int_1^2(1-x)dx\).

\(-1\)

\(D=\int_0^\pi\tfrac{\sin t}{1+\cos^2t}dt\) (\(x=\cos t\))

\(\int_{-1}^1\tfrac{dx}{1+x^2}\).

\(\tfrac\pi2\)

\(E=\int\tfrac{dt}{2t\ln t+t}\) (\(x=\ln t\))

\(\int\tfrac{dx}{2x+1}\) (bornes de l'énoncé ambiguës).

\(\tfrac12\ln|2\ln t+1|\)

\(F=\int_0^1\tfrac{dt}{1+e^t}\) (\(x=e^t\))

\(\int_1^e(\tfrac1x-\tfrac1{1+x})dx\).

\(1+\ln\tfrac2{1+e}\)

\(G=\int_1^3\tfrac{\sqrt t}{t+1}dt\) (\(x=\sqrt t\))

\(2\int_1^{\sqrt3}(1-\tfrac1{x^2+1})dx\).

\(2\sqrt3-2-\tfrac\pi6\)

(1) \(\int(\cos x)^{1234}\sin x\,dx\)

\(-\tfrac{(\cos x)^{1235}}{1235}\)

(2) \(\int\tfrac{dx}{\sin x}\)

\(\ln\big|\tan\tfrac x2\big|\)

(3) \(\int\tfrac1{x\ln x}dx\)

\(\ln|\ln x|\)

(4) \(\int\tfrac1{3+e^{-x}}dx\)

\(\times\tfrac{e^x}{e^x}\) : \(\tfrac{e^x}{3e^x+1}\).

\(\tfrac13\ln(3e^x+1)\)

(5) \(\int\tfrac1{\sqrt{4x-x^2}}dx\)

\(4x-x^2=4-(x-2)^2\).

\(\arcsin\tfrac{x-2}2\)

\(f_1=\tfrac{2t}{1+t^4}\) (\(u=t^2\))

\(\arctan(t^2)\)

\(f_2=\tfrac{e^{2t}}{1+e^t}\) (\(u=e^t\))

\(\int\tfrac u{1+u}du=u-\ln(1+u)\).

\(e^t-\ln(1+e^t)\)

\(f_3=\tfrac1{\operatorname{ch}t}\)

\(2\arctan(e^t)\)

\(f_4=\tfrac{\sin t\cos t}{\sin^2t+1}\) (\(u=\sin^2t+1\))

\(\tfrac12\ln(\sin^2t+1)\)

\(f_5=\cos(2\ln x)\)

\(\tfrac x5\big(2\sin(2\ln x)+\cos(2\ln x)\big)\)

\(f_6=\cos(\sqrt x)\)

\(2\big(\sqrt x\sin\sqrt x+\cos\sqrt x\big)\)

\(f_7=\tfrac{e^x}{(3+e^x)\sqrt{e^x-1}}\)

\(\arctan\big(\tfrac12\sqrt{e^x-1}\big)\)

\(f_8=\tfrac{e^x-1}{e^x+1}\)

\(2\ln(e^x+1)-x\)

\(A=\int_1^2\ln x\,dx\)

\(2\ln2-1\)

\(B=\int_1^2(\ln x)^2dx\)

\(2\ln^2 2-4\ln2+2\)

\(C=\int_0^1 xe^x\,dx\)

\(1\)

\(D=\int_1^e x^2\ln x\,dx\)

\(\tfrac{2e^3+1}9\)

\(E=\int_{\pi/6}^{\pi/3}e^t\sin(2t)dt\)

\(\tfrac15(e^{\pi/3}+e^{\pi/6})+\tfrac{\sqrt3}{10}(e^{\pi/3}-e^{\pi/6})\)

\(F=\int_0^{\pi/2}\cos t\,e^{-t}dt\)

\(\tfrac12(e^{-\pi/2}+1)\)

\(G=\int_{\pi/4}^{2\pi/3}t\sin t\,dt\)

\(\tfrac\pi3+\tfrac{\sqrt3}2+\tfrac1{\sqrt2}(\tfrac\pi4-1)\)

\(H=\int_0^1 t^2e^{-t}dt\)

\(-\tfrac5e+2\)

\(I=\int_1^2(x+1)e^{2x}dx\)

\(\tfrac14(5e^4-3e^2)\)

\(J=\int_0^1(2x^3+3x^2-x+1)e^x\,dx\)

\(4\)

\(K=\int_0^1 t\arctan t\,dt\)

\(\tfrac{\pi-2}4\)

(a) \(\arctan x\)

\(x\arctan x-\tfrac12\ln(x^2+1)\) (sur \(\mathbb R\))

(b) \(t\sin t\cos2t\)

\(\tfrac12 t\big(-\tfrac13\cos3t+\cos t\big)-\tfrac12\big(-\tfrac19\sin3t+\sin t\big)\)

(c) \(\sin(\ln x)\)

\(\tfrac12 x\sin(\ln x)-\tfrac12 x\cos(\ln x)\) (sur \(]0,+\infty[\))

\(f_1=\tfrac1{2x+1}\)

\(\tfrac12\ln|2x+1|\)

\(f_2=\tfrac{2x}{2x+1}=1-\tfrac1{2x+1}\)

\(x-\tfrac12\ln|2x+1|\)

\(f_3=\tfrac{3x+1}{2x+1}=\tfrac32-\tfrac{1/2}{2x+1}\)

\(\tfrac32x-\tfrac14\ln|2x+1|\)

\(f_4=\tfrac1{x^2+4}\)

\(\tfrac12\arctan\tfrac x2\)

\(f_5=\tfrac1{2x^2-3x-2}=\tfrac1{(2x+1)(x-2)}\)

\(-\tfrac15\ln|2x+1|+\tfrac15\ln|x-2|\)

\(f_6=\tfrac1{4x^2+4x+1}=\tfrac1{(2x+1)^2}\)

\(-\tfrac1{2(2x+1)}\)

\(f_7=\tfrac1{x^2-4x+5}=\tfrac1{(x-2)^2+1}\)

\(\arctan(x-2)\)

\(f_8=\tfrac1{x^2+4x+5}=\tfrac1{(x+2)^2+1}\)

\(\arctan(x+2)\)

\(f_9=\tfrac{3x+2}{x^2+x+1}\)

\(\tfrac32\ln(x^2+x+1)-\tfrac2{\sqrt3}\arctan\tfrac{2x+1}{\sqrt3}\)

\(f_{10}=\tfrac{2x}{x^2-x+1}\)

\(2x=(2x-1)+1\).

\(\ln(x^2-x+1)+\tfrac2{\sqrt3}\arctan\tfrac{2x-1}{\sqrt3}\)

\(f_{11}=\tfrac{2x+1}{x^2+x-3}\)

\(\ln|x^2+x-3|\)

(a) \(\int_2^3\tfrac{dx}{x(x+1)}\)

\(\tfrac1{x(x+1)}=\tfrac1x-\tfrac1{x+1}\).

\(\ln\tfrac98\)

(b) \(\int_0^{1/2}\tfrac{x+1}{(x^2+1)(x+2)}dx\)

\(=\tfrac{\frac15x+\frac35}{x^2+1}-\tfrac{1/5}{x+2}\).

\(-\tfrac1{10}\ln5+\tfrac35\arctan\tfrac12+\tfrac3{10}\ln2\)

(c) \(\int_4^5\tfrac{dx}{x^2-4x+3}\)

\(=(x-1)(x-3)\).

\(\tfrac12\ln\tfrac32\)

(d) \(\int_0^1\tfrac{dx}{1+x+x^2}\)

\((x+\tfrac12)^2+\tfrac34\).

\(\tfrac\pi{3\sqrt3}\)

(e) \(\int_3^4\tfrac{x^2-3x+1}{(x-1)^2(x-2)}dx\)

\(=-\tfrac1{x-2}+\tfrac2{x-1}+\tfrac1{(x-1)^2}\).

\(\ln\tfrac98+\tfrac16\)

(f) \(\int_0^1\tfrac{x^4-x^3+9x^2-13x+3}{x^3-2x^2+3x-6}dx\)

\(\tfrac32+2\ln2-\tfrac52\ln3\)

Éléments simples : \(f=\dfrac3{x-1}+\dfrac2{x+3}-\dfrac1{(x+3)^2}\). Primitive générale \(F=3\ln(x-1)+2\ln(x+3)+\tfrac1{x+3}+C\), condition \(F(2)=0\Rightarrow C=-2\ln5-\tfrac15\).

\(I=\int_1^2\tfrac{\ln(1+t)}{t^2}dt\)

IPP \(u=\ln(1+t),dv=\tfrac{dt}{t^2}\) puis \(\tfrac1{t(1+t)}=\tfrac1t-\tfrac1{1+t}\).

\(3\ln2-\tfrac32\ln3\)

\(J=\int_0^1 x(\arctan x)^2dx\)

\(\tfrac{\pi^2}{16}-\tfrac\pi4+\tfrac12\ln2\)

\(K=\int_a^1\tfrac{x\ln x}{(x^2+1)^2}dx\) (\(a>0\))

\(u=\ln x,\ dv=\tfrac x{(x^2+1)^2}dx,\ v=-\tfrac1{2(x^2+1)}\).

\(-\tfrac14\ln2+\tfrac14\ln(a^2+1)+\tfrac{\ln a}{2(a^2+1)}-\tfrac12\ln a\)

1) Valeurs

\(I(0,0)=1,\ I(1,0)=\tfrac12,\ I(1,1)=\tfrac16\)

2) Symétrie

Changement \(t\mapsto1-t\).

\(I(p,q)=I(q,p)\)

3) \(I(p,0)\)

\(\dfrac1{p+1}\)

4) Récurrence (IPP)

\(I(p,q+1)=\dfrac{q+1}{p+1}\,I(p+1,q)\)

5) \(I(p,1)\) et \(I(p,2)\)

\(I(p,1)=\tfrac1{p+1}I(p+1,0)=\tfrac1{(p+1)(p+2)}\) ; \(I(p,2)=\tfrac2{p+1}I(p+1,1)\).