Calculer les intégrales suivantes.
En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Intégrales 3Intégration en mathématiques/Exercices/Intégrales 3 », n'a pu être restituée correctement ci-dessus.
∫ 0 π 4 sin 3 x d x {\displaystyle \int _{0}^{\frac {\pi }{4}}\sin 3x\,\mathrm {d} x}
Solution
∫ 0 π 4 sin 3 x d x = − 1 3 [ cos 3 x ] 0 π 4 = 1 + 1 / 2 3 {\displaystyle \int _{0}^{\frac {\pi }{4}}\sin 3x\,\mathrm {d} x={\frac {-1}{3}}\left[\cos 3x\right]_{0}^{\frac {\pi }{4}}={\frac {1+1/{\sqrt {2}}}{3}}} .
∫ − π 4 π 4 tan 2 x d x {\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\tan ^{2}x\,\mathrm {d} x}
Solution
∫ − π 4 π 4 tan 2 x d x = [ tan x − x ] − π / 4 π / 4 = 2 − π 2 {\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\tan ^{2}x\,\mathrm {d} x=\left[\tan x-x\right]_{-\pi /4}^{\pi /4}=2-{\frac {\pi }{2}}} .
∫ − π 4 π 4 tan x cos x d x {\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}{\frac {\tan x}{\cos x}}\,\mathrm {d} x}
Solution
∫ − π 4 π 4 tan x cos x d x = 0 {\displaystyle \int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}{\frac {\tan x}{\cos x}}\,\mathrm {d} x=0} , par imparité.
∫ 0 π 3 cos 2 t sin 3 t d t {\displaystyle \int _{0}^{\frac {\pi }{3}}\cos 2t\sin ^{3}t\,\mathrm {d} t}
Solution
∫ 0 π 3 cos 2 t sin 3 t d t = ∫ 0 π 3 ( 2 cos 2 t − 1 ) ( cos 2 t − 1 ) ( − sin t ) d t = ∫ 1 1 / 2 ( 2 x 2 − 1 ) ( x 2 − 1 ) d x = ∫ 1 1 / 2 ( 2 x 4 − 3 x 2 + 1 ) d x = [ 2 x 5 5 − x 3 + x ] 1 1 / 2 = − 1 80 . {\displaystyle {\begin{aligned}\int _{0}^{\frac {\pi }{3}}\cos 2t\sin ^{3}t\,\mathrm {d} t&=\int _{0}^{\frac {\pi }{3}}\left(2\cos ^{2}t-1\right)\left(\cos ^{2}t-1\right)(-\sin t)\,\mathrm {d} t\\&=\int _{1}^{1/2}\left(2x^{2}-1\right)\left(x^{2}-1\right)\,\mathrm {d} x\\&=\int _{1}^{1/2}\left(2x^{4}-3x^{2}+1\right)\,\mathrm {d} x\\&=\left[{\frac {2x^{5}}{5}}-x^{3}+x\right]_{1}^{1/2}\\&={\frac {-1}{80}}.\end{aligned}}}
∫ 0 π 2 cos t cos 2 t cos 3 t d t {\displaystyle \int _{0}^{\frac {\pi }{2}}\cos t\cos 2t\cos 3t\,\mathrm {d} t}
Solution
2 cos a cos b = cos ( a + b ) + cos ( a − b ) {\displaystyle 2\cos a\cos b=\cos(a+b)+\cos(a-b)} donc
cos t cos 2 t cos 3 t = 1 2 ( cos 3 t + cos t ) cos 3 t = 1 4 ( 1 + cos 6 t + cos 4 t + cos 2 t ) {\displaystyle \cos t\cos 2t\cos 3t={\frac {1}{2}}(\cos 3t+\cos t)\cos 3t={\frac {1}{4}}(1+\cos 6t+\cos 4t+\cos 2t)} et
1 4 ∫ 0 π 2 ( 1 + cos 6 t + cos 4 t + cos 2 t ) d t = 1 4 [ t + sin 6 t 6 + sin 4 t 4 + sin 2 t 2 ] 0 π 2 = [ t 4 ] 0 π 2 = π 8 {\displaystyle {\frac {1}{4}}\int _{0}^{\frac {\pi }{2}}(1+\cos 6t+\cos 4t+\cos 2t)\,\mathrm {d} t={\frac {1}{4}}\left[t+{\frac {\sin 6t}{6}}+{\frac {\sin 4t}{4}}+{\frac {\sin 2t}{2}}\right]_{0}^{\frac {\pi }{2}}=\left[{\frac {t}{4}}\right]_{0}^{\frac {\pi }{2}}={\frac {\pi }{8}}} .
∫ π 4 5 π 4 ( cos 2 t + sin 4 t ) sin t d t {\displaystyle \int _{\frac {\pi }{4}}^{\frac {5\pi }{4}}\left(\cos ^{2}t+\sin ^{4}t\right)\sin t\,\mathrm {d} t}
Solution
∫ π 4 5 π 4 ( cos 2 t + sin 4 t ) sin t d t = − ∫ 1 / 2 − 1 / 2 ( x 2 + ( 1 − x 2 ) 2 ) d x = [ x 5 5 − x 3 3 + x ] − 1 / 2 1 / 2 = 2 ( 1 20 − 1 6 + 1 ) = 53 60 {\displaystyle \int _{\frac {\pi }{4}}^{\frac {5\pi }{4}}\left(\cos ^{2}t+\sin ^{4}t\right)\sin t\,\mathrm {d} t=-\int _{1/{\sqrt {2}}}^{-1/{\sqrt {2}}}\left(x^{2}+(1-x^{2})^{2}\right)\,\mathrm {d} x=\left[{\frac {x^{5}}{5}}-{\frac {x^{3}}{3}}+x\right]_{-1/{\sqrt {2}}}^{1/{\sqrt {2}}}={\sqrt {2}}\left({\frac {1}{20}}-{\frac {1}{6}}+1\right)={\frac {53}{60}}} .
∫ 0 π 2 cos t 5 − 3 sin t d t {\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\cos t}{5-3\sin t}}\,\mathrm {d} t}
Solution
∫ 0 π 2 cos t 5 − 3 sin t d t = ∫ 0 1 d x 5 − 3 x = − 1 3 [ ln ( 5 − 3 x ) ] 0 1 = ln ( 5 / 2 ) 3 {\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {\cos t}{5-3\sin t}}\,\mathrm {d} t=\int _{0}^{1}{\frac {\mathrm {d} x}{5-3x}}={\frac {-1}{3}}\left[\ln(5-3x)\right]_{0}^{1}={\frac {\ln(5/2)}{3}}} .
∫ π 6 π 3 2 sin t 1 + cos 2 t 2 d t {\displaystyle \int _{\frac {\pi }{6}}^{\frac {\pi }{3}}{\frac {2\sin t}{1+\cos ^{2}{\frac {t}{2}}}}\,\mathrm {d} t}
Solution
∫ π 6 π 3 2 sin t 1 + cos 2 t 2 d t = − 4 ∫ 7 / 4 3 / 4 d x x = 4 ln ( 7 / 3 ) {\displaystyle \int _{\frac {\pi }{6}}^{\frac {\pi }{3}}{\frac {2\sin t}{1+\cos ^{2}{\frac {t}{2}}}}\,\mathrm {d} t=-4\int _{7/4}^{3/4}{\frac {\mathrm {d} x}{x}}=4\ln(7/3)} .