Correction exercice

Calculons les intégrales suivantes :
a) \(I = \) \(\int {\frac{{{{\sin }^3}x}}{{2 + \cos x}}} dx\) \( = \) \(\int {\frac{{{{\sin }^2}x}}{{2 + \cos x}}} \sin xdx\) \( = \) \(\int {\frac{{\left( {1 - {{\cos }^2}x} \right)}}{{2 + \cos x}}} \) \(\sin xdx\)
Effectuons le changement de variable \(\cos x = t\), alors \(\sin xdx = - dt\)
\(I = \) \(\int {\frac{{{t^2} - 1}}{{2 + t}}} dt = \) \(\int {(t - 2 + \frac{3}{{2 + t}}} \) \()dt = \frac{{{{\cos }^2}x}}{2}\) \( - 2\cos x + \) \(Log(\cos x + 2)\) \( + cte\)

b) \(I = \) \(\int {\frac{{dx}}{{2 - {{\sin }^2}x}}} \)
Posons \(\tan x = t\)
\(I = \) \(\int {\frac{1}{{\left( {2 - \frac{{{t^2}}}{{1 + {t^2}}}} \right)}}} \) \(\frac{{dt}}{{1 + {t^2}}} = \) \(\int {\frac{{dt}}{{2 + {t^2}}}} \) \( = \frac{1}{{\sqrt 2 }}\arctan \frac{t}{{\sqrt 2 }}\) \( = \frac{1}{{\sqrt 2 }}\) \(\arctan \left( {\frac{{\tan x}}{{\sqrt 2 }}} \right)\) \( + cte\)

c) \(I = \int {\frac{{{{\cos }^3}x}}{{{{\sin }^4}x}}} dx\) \( = \) \(\int {\frac{{{{\cos }^2}x\cos x}}{{{{\sin }^4}x}}} dx\)
\(I = \) \(\int {\frac{{\left( {1 - {{\sin }^2}x} \right)\cos x}}{{{{\sin }^4}x}}} \) \(dx\)
Posons \(\sin x = t\)
\(I = \) \(\int {\frac{{\left( {1 - {t^2}} \right)}}{{{t^4}}}} dt\) \( = \int {\frac{{dt}}{{{t^4}}}} - \) \(\int {\frac{{dt}}{{{t^2}}}} = - \) \(\frac{1}{{3{t^3}}} + \frac{1}{t} = \) \( - \frac{1}{{3{{\sin }^3}x}} + \) \(\frac{1}{{\sin x}} + cte\)

d) \(I = \) \(\int {\frac{{{{\sin }^2}x}}{{{{\cos }^6}x}}dx} \) \( = \) \(\int {\frac{{{{\sin }^2}x\left( {{{\cos }^2}x + {{\sin }^2}x} \right)}}{{{{\cos }^6}x}}dx} \) \( = \int {{{\tan }^2}x} \) \({\left( {1 + {{\tan }^2}x} \right)^2}dx\)
Posons \(\tan x = t\)
\(I = \int {{t^2}} {\left( {1 + {t^2}} \right)^2}\) \(\frac{{dt}}{{1 + {t^2}}}\)
\(I = \frac{{{{\tan }^3}x}}{3}\) \( + \frac{{{{\tan }^5}x}}{5} + cte\)

e) \(I = \int {\cos mx} \) \(\cos nxdx\)
\(I = \int {\cos mx} \) \(\cos nxdx = \) \(\frac{1}{2}\) \(\int {(\cos (m + n)x} \) \( + \cos (m - n)x)dx\)
\(I = \frac{{\sin (m + n)x}}{{2(m + n)}}\) \( + \frac{{\sin (m - n)x}}{{2(m - n)}}\) \( + cte\)
f) \(I = \int {\sin 5x} \) \(\sin 3xdx = \) \(\frac{1}{2}\int {( - \cos 8x} \) \( + \cos 2x)dx = \) \( - \frac{{\sin 8x}}{{12}} + \) \(\frac{{\sin 2x}}{4} + cte\)

g) \(I = \int {\frac{{dx}}{{{{\cos }^4}x}}} \) \( = \tan x + \) \(\frac{1}{3}{\tan ^3}x + cte\)