Title: Integration composee
Course: Prérequis de mathématiques (Mathuselem)

Q: Calculer l’intégrale suivante : \(I = \int_{0}^{1} \ t\ e^{2t^2} \ dt\).
- \(I=1-e\).
- \(I=\displaystyle \frac12 (e-1)\).
- \(I=\displaystyle \frac14 (e^4-1)\).
- \(I=2e^2-1\).
A: \(I=\displaystyle \frac14 (e^4-1)\).
Feedback: On pose : \(u(t) = t^2\) et \(u'(t) = 2t.\) \[ \begin{eqnarray*} I & = & \int_{0}^{1} \frac{1}{4} u'(t)e^{u(t)} \ dt\\ &= &\frac{1}{4} \ \left[ e^{2t^2}\right]_{0} ^{1} \\ & = & \dfrac{1}{4} \ \left(e^{2} +1\right). \end{eqnarray*} \]

Q: Calculer l’intégrale suivante \({I = \int_{1}^{e} \frac {\ln^2(t)} t \ dt}\) .
- \(I=\displaystyle\frac 12 (e-1)\).
- \(I= 2e-1\).
- \(I=\displaystyle\frac 14 (e^2-1)\).
- \(I=\displaystyle \frac 13.\)
A: \(I=\displaystyle \frac 13.\)
Feedback: On pose : \(u(t) = \ln t\) et \(u'(t) = \displaystyle\frac{1}{t}.\) \[ \begin{eqnarray*} I & = & \int_{1}^{e} \ \frac{1}{3} 3u^2(t)u'(t) \ dt\\ &= & \left[ \frac 13 u(t)^{2} \right]_{1}^{e} \\ &= & \left[ \frac 13 \ln(t)^{2} \right]_{1}^{e} \\ & = & \frac{1}{3}. \end{eqnarray*} \]

Q: Calculer l’intégrale suivante :$I = _{0}^{} \sin t^{2}t   dt $.
- \(I = 0\).
- \(I = 2\).
- \(I =\frac{1}{2}\).
- \(I=\frac{1}{3}\)
- \(I=1\)
A: \(I = 2\).
Feedback: On pose \(u(x)=\cos x\), donc \(u'(x)=\ -\ \sin x\). \[ \begin{eqnarray*} I & = & -\frac 13{ \int_{0}^{\pi}3 u'(t)u^{2}(t) \ dt}\\ &= & 0-\frac{1}{3} \ \left[ u^3(t) \right]_{0} ^{\pi} \\ & = &- \frac{1}{3} \left( \cos^3\pi - \cos^0 0 \right) \\ & = & 2. \end{eqnarray*} \]

Q: Calculer l’intégrale suivante :\(I = \displaystyle\int_{0}^{1} \ \frac{e^t-e^{-t}} {1+ e^{t}+e^{-t}} \ dt\).
- \(I = \ln(1+e) -\ln(2)\).
- \(I = \ln (1+e+e^{-1})-\ln(3)\).
- \(I = \ln(2+e) -1\).
- \(I = \ln(2-e+e^{-1}) -2\).
A: \(I = \ln (1+e+e^{-1})-\ln(3)\).
Feedback: On pose \(u(x) =1+ e^x+e^{-x}\). On a \(u'(x) = e^x-e^{-x}\). \[ \begin{eqnarray*} I & = & \int_{0}^{1}\frac {u'(t)} {1+u(t)}\ dt\\ &= &\left[ \ln (u(t)) \right]_{0}^{1}\\ & = & \left[\ln(1+e^t+e^{-t}) \right]_{0}^{1}\\ & = &\ln (1+e+e^{-1})-\ln(3). \end{eqnarray*} \]

Q: Calculer l’intégrale suivante : \({I = \int_{1}^{3} \ \frac{ 1+ \sqrt{t}+t}{\sqrt{t}} \ dx}\).
- \(I = 4\sqrt{3}-\frac 23\).
- $I = 2 - $.
- \(I = 3 \sqrt{3} - \sqrt{2}\).
- \(I = 4\sqrt{3}-\frac 43\).
A: \(I = 4\sqrt{3}-\frac 23\).
Feedback: On pose \(u(x) = \sqrt{x}\) et on a \(u'(x) =\frac{1}{2\sqrt{x}}.\)$ \[ \begin{eqnarray*} I & = & 2\int_{1}^{3} \ (1+u(t)+u^2(t))u'(t) \ dt\\ &= & 2\left[ u(t)\right]_{1}^{3} +\left[ u^2(t)\right]_{1}^{3}+ \frac 23 \left[ u^3(t)\right]_{1}^{3}\\ &= & 2\left[ \sqrt{t}\right]_{1}^{3} +\left[ t\right]_{1}^{3}+ \frac 23 \left[ t\sqrt{t}\right]_{1}^{3}\\ & = & 2(\sqrt{3}-1)+ 3-1 +\frac 23 (3\sqrt{3}-1) \\ & = & 4\sqrt{3}-\frac 23. \end{eqnarray*} \]

Q: Calculer l’intégrale suivante :\(\displaystyle I = \int_{\pi/3}^{\pi/2} \frac{ \cos t}{\sin t}\ dt\).
- \(I =\displaystyle \frac \pi 2 - \frac {4\pi} 3\).
- \(I = \displaystyle\ln (2) -\frac 12 \ln (3)\).
- \(I = \displaystyle\frac 13 (\ln (2)-1)\).
- \(I = \displaystyle\frac 12 (\ln (3)-\ln(2))\).
A: \(I = \displaystyle\ln (2) -\frac 12 \ln (3)\).
Feedback: On pose \(u(x) =\sin x\), donc \(u'(x) = \cos x\). \[ \begin{eqnarray*} I & = & \int_{\pi/3}^{\pi/2} \ \frac{ u'(t)}{u(t)} \ dt\\ &=&\left[ \ln(u(t))\right]_{\pi/3}^{\pi/2} \\ &=& \left[ \ln(\sin t)\right]_{\pi/3}^{\pi/2} \\ & = & \ln \left(\frac 1{\sqrt 2}\right)- \ln \left(\frac{\sqrt 3} 2\right) \\ &=& 0-\frac 12 \ln (3)+\ln(2)\\ &=& \ln (2) -\frac 12 \ln (3). \end{eqnarray*} \]

Q: Calculer l’intégrale suivante :\(\displaystyle I = \int_{0}^{\pi/2} \cos^3t \ dt\).
- \(I =\displaystyle \frac 14\).
- \(I = \displaystyle\frac 23\).
- \(I = \displaystyle\frac 12\).
- \(I = \displaystyle\frac{ 2}{\sqrt 3}\).
- \(I = 1\).
A: \(I = \displaystyle\frac 23\).
Feedback: Contrairement à ce qu’on attend, on pose \(u(x) =\sin x\), donc \(u'(x) = \cos x\). Il va falloir transformer la fonction \(\cos^2\) en \(\sin^2\), ce qui est possible car l’exposant est pair : \(\cos^2x=1-\sin^2x\) \[\begin{eqnarray*} I & = &\int_{0}^{\pi/2} \cos^3t \ dt\\ & = & \int_{0}^{\pi/2} \cos^2t \cos t\ dt\\ & = & \int_{0}^{\pi/2} (1-\sin^2t) \cos t\ dt\\ & = &\int u'(t)- \frac 13(3u^2(t)u'(t))\ dt\\ &=& \left[u(t)-\frac 13u^3(t)\right]_{0}^{\pi/2}\\ &=& \left[\sin t-\frac 13\sin^3t\right]_{0}^{\pi/2}\\ &=& 1-\frac 13-0\\ &=& \frac 23. \end{eqnarray*}\]

Q: Calculer l’intégrale suivante :\(\displaystyle I = \int_{\pi/3}^{\pi/2} \frac{ \cos t}{\sin^2 t}\ dt\).
- \(I =\displaystyle -\frac{ 2}{\sqrt 3}+\sqrt 2\).
- \(I = \displaystyle\frac 12 \left(\frac{ 2}{\sqrt 3}-1\right)\).
- \(I = \displaystyle\frac{ 2}{\sqrt 3}-\sqrt 2\).
- \(I = \displaystyle\frac 12 (\ln (3)-1)\).
A: \(I = \displaystyle\frac 12 \left(\frac{ 2}{\sqrt 3}-1\right)\).
Feedback: On pose \(u(x) =\sin x\), donc \(u'(x) = \cos x\). \[\begin{eqnarray*} I & = &- \int_{\pi/3}^{\pi/2} \ - \frac{ u'(t)}{u^2(t)} \ dt\\ &=&\left[ \frac{ 1}{u(t)} ]_{\pi/3}^{\pi/2} \\ &=& left[ \frac{ 1}{\sin t}]_{\pi/3}^{\pi/2} \\ & = & \frac{ 2}{\sqrt 3}-\sqrt 2. \end{eqnarray*}\]

Q: Calculer l’intégrale suivante : \(I =\displaystyle \int_{0}^{\pi/2} \frac {\sin t}{\sqrt{\cos t}} \ dt\).
- \(I = 1.\)
- \(I = 2.\)
- \(I = \displaystyle\frac{2}{3}.\)
- $I =2☻ $.
A: \(I = 2.\)
Feedback: On pose \(u(x) =\cos x\), donc \(u'(x) =- \sin x\). On reconnait la dérivée de \(\sqrt{u(x)}\). \[ \begin{eqnarray*} I & = & -2 \displaystyle \int_{0}^{\pi/2} \frac {u'(t)}{2\sqrt{u(t)}} \ dt & = & -2 \left[ \sqrt{u(t)} \right]_{0}^{\pi/2} \\ & = & -2 \left[\sqrt{ \cos t }\right]_{0}^{\pi/2} \\ & = & -2(0-1)\\ &=&2. \end{eqnarray*} \]

Q: Calculer l’intégrale suivante : \({I = \displaystyle\int_{0}^{1} \frac {3t^2+2}{t^3+2t+1} \ dt}\).
- \(I = 1.\)
- \(I = 15.\)
- \(I = \displaystyle\frac{2}{3}.\)
- \(I=\ln(4)\).
- \(I =\sqrt 3\).
A: \(I=\ln(4)\).
Feedback: La fraction rationelle est continue et bornée sur l’intervalle \([0;1]\) donc intégrable sur cet intervalle. On pose \(u(x) = x^3+2x+1\), donc \(u'(x) =3x^2+2\). \[\begin{eqnarray*} I & = & \displaystyle \int_{0}^{1} \frac {u'(t)}{u(t)} \ dt\\ & = & \left[ \ln(u(t)) \right]_{0}^{1} \\ & = & \left[\ln(t^3+2t+1) \right]_{0}^{1} \\ & = & \ln(4)-\ln(1)\\ &=& \ln(4). \end{eqnarray*}\]

Q: Calculer l’intégrale suivante : \(I =\displaystyle \int_{0}^{\pi^2} \frac{\cos (\sqrt{ t})} {\sqrt{ t}}\ dt\).
- \(I = 3.\)
- \(I = 0.\)
- \(I = \displaystyle\frac{2}{3}.\)
- \(I =\sqrt \pi\).
A: \(I = 0.\)
Feedback: On pose \(u(x) =\sqrt{ t}\), donc \(u'(x) =\displaystyle 1 {2\sqrt{ t}}\). \[ \begin{eqnarray*} I & = & 2 \displaystyle \int_{0}^{\pi^2} \cos (u(t)) u'(t) \ dt\\ & = & 2\ \left[ \sin (u(t)) \right]_{0}^{\pi^2} \\ & = & 2 \ \left[\sin(\sqrt{ t })\right]_{0}^{\pi^2} \\ & = & 0 \end{eqnarray*} \]

Q: Calculer l’intégrale suivante : \(I =\displaystyle \int_{0}^{3} \frac{(t+2)^2} {\sqrt{ t+1}}\ dt\).
- \(I = 3.\)
- \(I = 1.\)
- \(I = \displaystyle\frac{2}{3}.\)
- \(I = \displaystyle \frac {19}3\).
A: \(I = \displaystyle\frac{2}{3}.\)
Feedback: On pose \(u(x) =\sqrt{ t+1}\), donc \(u'(x) =\displaystyle \frac 1 {2\sqrt{ t+1}}\). On note que \((t+2)^2=(t+1)^2+2(t+1)+1=u^2(t)+2u(t)+1\). \[ \begin{eqnarray*} I & = & 2 \int_{0}^{4} ( u^2(t)+2u(t)+1) u'(t) \ dt\\ & = & 2\left[ \frac 13 u^3(t)+u^2(t)+u(t) \right]_{0}^{3} \\ & = & 2 \ \left[\frac 13 (t+1)\sqrt{t+1} +t+1+\sqrt {t+1} \right]_{0}^{3} \\ & = & \frac 83-\frac 13+4-1+2-1\\ &=&\frac {19}3. \end{eqnarray*} \]

Q: Calculer l’intégrale suivante : \(I =\displaystyle \int_{0}^{\pi/4} \frac{2\sin t\cos t} {\cos^2 t+1}\ dt\).
- \(I = 1.\)
- \(I = 2\ln(2)-\ln(3).\)
- \(I = \displaystyle\frac{1}{3}.\)
- \(I = \ln(2)+\ln(3)\).
A: \(I = 2\ln(2)-\ln(3).\)
Feedback: On pose \(u(x) =\cos^2 x\), donc \(u'(x) =-2\sin x\cos x\). \[ \begin{eqnarray*} I & = &-\int_{0}^{\pi/4} \frac{u'(t)}{u(t)+1} \ dt\\ & = & -\ \left[ \ln (u(t)+1) \right]_{0}^{\pi/4} \\ & = &- \ \left[\ln(\cos^2 t+1)\right]_{0}^{\pi/4} \\ & = & -\ln\left(\frac 32\right)+ \ln\left(2\right)\\ &=& 2\ln(2)-\ln(3). \end{eqnarray*} \]

Q: Calculer l’intégrale suivante : \(I =\displaystyle \int_{0}^{2} \frac{t+2} {(t^2+4 t+3)}\ dt\).
- \(I = 4.\)
- \(I = \ln(2)-\ln(12).\)
- \(I = \displaystyle\frac 2{15}.\)
- \(I = \sqrt 2\).
- \(I = \displaystyle\frac 35.\)
A: \(I = \displaystyle\frac 2{15}.\)
Feedback: On pose \(u(x) =x^2+4x+3\), donc \(u'(x) =2x+4\). $. \[ \begin{eqnarray*} I & = &-\frac 12\int_{0}^{2} \frac{-u'(t)}{u^2(t)} \ dt\\ & = & -\frac 12\ \left[ \frac 1{u(t)} \right]_{0}^{2} \\ & = &-\frac 12 \ \left[\frac1{t^2+4t+3} \right]_{0}^{2} \\ & = & -\frac 12\left(\frac 1{15} -\frac 13\right)\\ &=& \frac 2{15}. \end{eqnarray*} \]

Q: Calculer l’intégrale suivante :\(\displaystyle I = \int_{\pi/4}^{\pi/3} \frac{ \sin t}{\cos t}\ dt\).
- \(I =\displaystyle \frac \pi 2 - \frac {4\pi} 3\).
- \(I = \displaystyle\frac 12 (\ln (2)+\ln (3))\).
- \(I = \displaystyle\frac 12 \ln (2)\).
- \(I = \displaystyle\frac 12 (\ln (3)-\ln(2))\).
- \(I = 1\).
A: \(I = \displaystyle\frac 12 \ln (2)\).
Feedback: On pose \(u(x) =\cos x\), donc \(u'(x) = -\sin x\). \[ \begin{eqnarray*} I & = & -\int_{\pi/4}^{\pi/3} \ \frac{ u'(t)}{u(t)} \ dt\\ &=&-\left[ \ln(u(t))\right]_{\pi/4}^{\pi/3} \\ &=&- \left[ \ln(\cos t)\right]_{\pi/4}^{\pi/3} \\ & = &- \ln \left(\frac 12\right)+ \ln \left(\frac 1{\sqrt 2}\right) \\ &=& \ln (2)-\frac 12 \ln(2)\\ &=& \frac 12 \ln (2). \end{eqnarray*} \]