A la limite du possible ! - Exercice 1

Notons par $$I$$ l'intervalle $$\left] 0 \,;\, 2 \right]$$.
Sur cet intervalle la fonction $$t \longmapsto \dfrac{t}{\tan^2(t)}$$ est continue, et donc la limite $$\lim_{x \, \longrightarrow \, 0} \int_x^{3x} \dfrac{t}{\tan^2(t)} \, dt$$ existe bien.
On a :
$$\int_x^{3x} \dfrac{t}{\tan^2(t)} \, dt = \int_x^{3x} \dfrac{t \cos^2(t)}{\sin^2(t)} \, dt = \int_x^{3x} \dfrac{t \big( 1 - \sin^2(t) \big)}{\sin^2(t)} \, dt = \int_x^{3x} \dfrac{t - t\sin^2(t) \big)}{\sin^2(t)} \, dt$$
Ce qui nous donne :
$$\int_x^{3x} \dfrac{t}{\tan^2(t)} \, dt = \int_x^{3x} \left( \dfrac{t}{\sin^2(t)} - \dfrac{ t\sin^2(t)}{\sin^2(t)} \right) \, dt = \int_x^{3x} \left( \dfrac{t}{\sin^2(t)} - t \right) \, dt$$
Par linéarité :
$$\int_x^{3x} \dfrac{t}{\tan^2(t)} \, dt = \int_x^{3x} \dfrac{t}{\sin^2(t)} \, dt - \int_x^{3x} t \, dt$$
Avec :
$$\int_x^{3x} t \, dt = \left[ \dfrac{t^2}{2} \right]_x^{3x} = \dfrac{(3x)^2}{2} - \dfrac{x^2}{2} = \dfrac{9x^2}{2} - \dfrac{x^2}{2} = \dfrac{9x^2 - x^2}{2} = \dfrac{8x^2}{2} = 4x^2$$
Puis, à l'aide d'une intégration par parties, et en se souvenant que $$\left( -\dfrac{1}{\tan(t)}\right)' = \dfrac{1}{\sin^2(t)}$$, on a :
$$\int_x^{3x} \dfrac{t}{\sin^2(t)} \, dt = \int_x^{3x} t \times \dfrac{1}{\sin^2(t)} \, dt = \left[ t \times \left( -\dfrac{1}{\tan(t)}\right) \right]_x^{3x} - \int_x^{3x} 1 \times \dfrac{-1}{\tan(t)} \, dt$$
Soit :
$$\int_x^{3x} \dfrac{t}{\sin^2(t)} \, dt = \left[ -\dfrac{t}{\tan(t)} \right]_x^{3x} + \int_x^{3x} \dfrac{1}{\tan(t)} \, dt = \left[ \dfrac{t}{\tan(t)} \right]_{3x}^{x} + \int_x^{3x} \dfrac{\cos(t)}{\sin(t)} \, dt$$
Soit encore :
$$\int_x^{3x} \dfrac{t}{\sin^2(t)} \, dt = \dfrac{x}{\tan(x)} - \dfrac{3x}{\tan(3x)} + \int_x^{3x} \dfrac{\big(\sin(t)\big)'}{\sin(t)} \, dt = \dfrac{x}{\tan(x)} - \dfrac{3x}{\tan(3x)} + \left[ \ln \left( | \sin(t)| \right) \right]_x^{3x}$$
Avec :
$$\left[ \ln \left( | \sin(t)| \right) \right]_x^{3x} = \ln \left( | \sin(3x)| \right) - \ln \left( | \sin(x)| \right) = \ln \left( \dfrac{| \sin(3x)|}{| \sin(x)|} \right) = \ln \left( \left| \dfrac{ \sin(3x)}{\sin(x)} \right|\right)$$
Donc :
$$\int_x^{3x} \dfrac{t}{\sin^2(t)} \, dt = \dfrac{x}{\tan(x)} - \dfrac{3x}{\tan(3x)} + \ln \left( \left| \dfrac{ \sin(3x)}{\sin(x)} \right|\right)$$
Ainsi, on a :
$$\int_x^{3x} \dfrac{t}{\tan^2(t)} \, dt = \dfrac{x}{\tan(x)} - \dfrac{3x}{\tan(3x)} + \ln \left( \left| \dfrac{ \sin(3x)}{\sin(x)} \right|\right) - 4x^2$$
D'où :
$$\lim_{x \, \longrightarrow \, 0} \int_x^{3x} \dfrac{t}{\tan^2(t)} \, dt = \lim_{x \, \longrightarrow \, 0} \left( \dfrac{x}{\tan(x)} - \dfrac{3x}{\tan(3x)} + \ln \left( \left| \dfrac{ \sin(3x)}{\sin(x)} \right|\right) - 4x^2 \right)$$
Ce qui nous donne :
$$\lim_{x \, \longrightarrow \, 0} \int_x^{3x} \dfrac{t}{\tan^2(t)} \, dt = \lim_{x \, \longrightarrow \, 0} \dfrac{x}{\tan(x)} - \lim_{x \, \longrightarrow \, 0} \dfrac{3x}{\tan(3x)} + \lim_{x \, \longrightarrow \, 0} \ln \left( \left| \dfrac{ \sin(3x)}{\sin(x)} \right|\right) - \lim_{x \, \longrightarrow \, 0} 4x^2$$
Et on a, puisque $$x \, \longrightarrow \, 0$$, $$\tan(x) \underset{0}{\sim} x$$ ainsi que $$\sin(x) \underset{0}{\sim} x$$ :
$$\bullet \,\, \lim_{x \, \longrightarrow \, 0} \dfrac{x}{\tan(x)} = \lim_{x \, \longrightarrow \, 0} \dfrac{x}{x} = \lim_{x \, \longrightarrow \, 0} \dfrac{1}{1} = \lim_{x \, \longrightarrow \, 0} 1 = 1$$
$$\bullet \bullet \,\, \lim_{x \, \longrightarrow \, 0} \dfrac{3x}{\tan(3x)} = \lim_{x \, \longrightarrow \, 0} \dfrac{3x}{3x} = \lim_{x \, \longrightarrow \, 0} \dfrac{x}{x} = \lim_{x \, \longrightarrow \, 0} \dfrac{1}{1} = \lim_{x \, \longrightarrow \, 0} 1 = 1$$
$$\bullet \bullet \bullet \,\, \lim_{x \, \longrightarrow \, 0} \ln \left( \left| \dfrac{ \sin(3x)}{\sin(x)} \right|\right) = \lim_{x \, \longrightarrow \, 0} \ln \left( \left| \dfrac{3x}{x} \right|\right) = \lim_{x \, \longrightarrow \, 0} \ln \left( \left| 3 \right|\right) = \lim_{x \, \longrightarrow \, 0} \ln \left( 3 \right) = \ln \left( 3 \right)$$
$$\bullet \bullet \bullet \bullet \,\, \lim_{x \, \longrightarrow \, 0} 4x^2 = 4 \lim_{x \, \longrightarrow \, 0} x^2 = 0$$
Ce qui nous donne :
$$\lim_{x \, \longrightarrow \, 0} \int_x^{3x} \dfrac{t}{\tan^2(t)} \, dt = 1 - 1 + \ln \left( 3 \right) - 0$$
Finalement, on obtient :
$$\lim_{x \, \longrightarrow \, 0} \int_x^{3x} \dfrac{t}{\tan^2(t)} \, dt = \ln \left( 3 \right)$$