Changement de variables - Exercice 2

Soit $$x\in \left[0,\frac{\pi }{6}\right]$$, on a : $$f\left(x\right)=\frac{1}{{\left({\mathrm{cos} \left(x\right)\ }\right)}^6}$$
On pose $$\red{t={\mathrm{tan} \left(x\right)\ }}$$ ce qui donne $$dt=\left(1+{\left({\mathrm{tan} \left(x\right)\ }\right)}^2\right)dx\Longrightarrow dt=\left(1+t^2\right)dx\Longrightarrow {\color{blue}{\frac{dt}{\left(1+t^2\right)}=dx}}$$
$$\int{\frac{1}{{\left({\mathrm{cos} \left(x\right)\ }\right)}^6}dx}=\int{{\left(\frac{1}{{\left({\mathrm{cos} \left(x\right)\ }\right)}^2}\right)}^3dx}$$
$$\int{\frac{1}{{\left({\mathrm{cos} \left(x\right)\ }\right)}^6}dx}=\int{{\left(1+{\left(\red{{\mathrm{tan} \left(x\right)\ }}\right)}^2\right)}^3{\color{blue}{dx}}}$$
$$\int{\frac{1}{{\left({\mathrm{cos} \left(x\right)\ }\right)}^6}dx}=\int{{\left(1+\red{t}^2\right)}^3}\times {\color{blue}{\frac{dt}{\left(1+t^2\right)}}}$$
$$\int{\frac{1}{{\left({\mathrm{cos} \left(x\right)\ }\right)}^6}dx}=\int{{\left(1+t^2\right)}^2}dt$$
$$\int{\frac{1}{{\left({\mathrm{cos} \left(x\right)\ }\right)}^6}dx}=\int{\left(1+2t^2+t^4\right)}dt$$
$$\int{\frac{1}{{\left({\mathrm{cos} \left(x\right)\ }\right)}^6}dx}=t+\frac{2}{3}t^3+\frac{1}{5}t^5+K$$ où $$K \in \mathbb{R}$$
Finalement :

Soit $$t\in \left[-1,+\infty\right[$$, on a : $$f\left(t\right)=\frac{1}{\sqrt{1+t}+3}$$ .
On pose $$\red{u=\sqrt{1+t}}\Longrightarrow u^2=1+t$$ ainsi $${\color{blue}{2udu=dt}}$$
$$\int{\frac{1}{\red{\sqrt{1+t}}+3}{\color{blue}{dt}}}=\int{\frac{1}{\red{u}+3}{\color{blue}{2udu}}}$$
$$\int{\frac{1}{\sqrt{1+t}+3}dt}=2\int{\frac{u}{u+3}du}$$
$$\int{\frac{1}{\sqrt{1+t}+3}dt}=2\int{\frac{u+3-3}{u+3}du}$$
$$\int{\frac{1}{\sqrt{1+t}+3}dt}=2\int{\frac{u+3}{u+3}du}+\ 2\int{\frac{-3}{u+3}du}$$
$$\int{\frac{1}{\sqrt{1+t}+3}dt}=2\int{1du}-6\int{\frac{1}{u+3}du}$$
$$\int{\frac{1}{\sqrt{1+t}+3}dt}=2u-6{\mathrm{ln} \left|u+3\right|\ }$$
$$\int{\frac{1}{\sqrt{1+t}+3}dt}=2\sqrt{1+t}-6{\mathrm{ln} \left|\sqrt{1+t}+3\right|\ }+K$$ où $$K \in \mathbb{R}$$
Ainsi :

Soit $$t\in \left[0;\frac{\pi }{3}\right]$$
$$\int{f\left(t\right)}dt=\int{\frac{t^2}{\sqrt{1-t^2}}dt}$$
On pose : $$\red{t={\mathrm{sin} \left(x\right)\ }}$$ ainsi $${\color{blue}{dt={\mathrm{cos} \left(x\right)dx\ }}}$$
D'où :
$$\int{f\left(t\right)}dt=\int{\frac{{\sin\ }^2\left(x\right)}{\sqrt{1-{\sin\ }^2\left(x\right)}}{\mathrm{cos} \left(x\right)dx\ }}$$
$$\int{f\left(t\right)}dt=\int{\frac{{\sin\ }^2\left(x\right)}{\sqrt{{\cos\ }^2\left(x\right)}}{\mathrm{cos} \left(x\right)dx\ }}$$
$$\int{f\left(t\right)}dt=\int{\frac{{\sin\ }^2\left(x\right)}{\left|{\mathrm{cos} \left(x\right)\ }\right|}{\mathrm{cos} \left(x\right)dx\ }}$$
Comme $$x\in \left[0;\frac{\pi }{3}\right]\$$alors $${\mathrm{cos} \left(x\right)\ }>0$$, il vient alors que :
$$\int{f\left(t\right)}dt=\int{\frac{{\sin\ }^2\left(x\right)}{{\mathrm{cos} \left(x\right)\ }}{\mathrm{cos} \left(x\right)dx\ }}$$
$$\int{f\left(t\right)}dt=\int{{\sin\ }^2\left(x\right)}dx$$
$$\int{f\left(t\right)}dt=\int{\frac{1}{2}\left(1-{\mathrm{cos} \left(2x\right)\ }\right)}dx$$
$$\int{f\left(t\right)}dt=\frac{1}{2}x-\frac{1}{4}{\mathrm{sin} \left(2x\right)\ }+K$$ où $$K \in \mathbb{R}$$
Or $$t={\mathrm{sin} \left(x\right)\ }$$ alors $${\mathrm{arcsin} \left(t\right)\ }=x$$, ce qui donne :

Finalement :

$$\int{f\left(t\right)}dt=\int{\frac{dt}{e^t-e^{-t}}}$$
On pose $$\red{x=e^t}$$ ainsi $$dx=e^tdt\Longrightarrow$$ $$\frac{dx}{e^t}=dt\Longrightarrow {\color{blue}{\frac{dx}{x}=dt}}$$
Il vient alors :
$$\int{f\left(t\right)}dt=\int{\frac{dt}{e^t-\frac{1}{e^t}}}$$
$$\int{f\left(t\right)}dt=\int{\frac{\frac{dx}{x}}{x-\frac{1}{x}}}$$
$$\int{f\left(t\right)}dt=\int{\frac{\frac{dx}{x}}{\frac{x^2-1}{x}}}$$
$$\int{f\left(t\right)}dt=\int{\frac{dx}{x^2-1}}$$
$$\int{f\left(t\right)}dt=-\int{\frac{dx}{1-x^2}}$$
$$\int{f\left(t\right)}dt=\frac{1}{2}{\mathrm{ln} \left|\frac{1+x}{1-x}\right|\ }+K$$ où $$K \in \mathbb{R}$$
Or $$x=e^t$$ ce qui nous donne :