Integrale de la juantheron

[juantheron]

[tex]\displaystyle \int\frac{x}{\sqrt{(7x-10-x^2)^3}}dx[/tex]

[HarapAlb]

Vad ca te pricepi la scris formule in LaTeX.

[florin_try]

Vad ca te pricepi la scris formule in LaTeX.

Nu are limite de integrare.

[zec]

O integrala dificila totusi:
Inainte de a incepe trebuie remarcat faptul ca domeniul maxim e intervalul (2,5).
Avem[tex]\int\frac{x}{(\sqrt{7x-10-x^2})^3}dx=-1/2\int\frac{-7+7-2x}{(\sqrt{7x-10-x^2})^3}dx=7/2\int\frac{1}{(\sqrt{7x-10-x^2})^3}dx+(-1/2)\int\frac{(7x-10-x^2)'}{(\sqrt{7x-10-x^2})^3}dx=(7x-10-x^2)^{-1/2}+7/2\int\frac{dx}{(\sqrt{(9/4-(x-7/2)^2)})^3[/tex]
Separat rezolvam [tex]\int\frac{dx}{(\sqrt{(9/4-(x-7/2)^2)})^3[/tex]
Se face substitutia x-7/2=9/4siny avem dx=9/4cosydy si integrala devine:
[tex]\int\frac{9/4cosy}{27/8cos^3y}dy=2/3\int\frac{dy}{cos^2y}=2/3tgy+C=2/3\frac{x-7/2}{\sqrt{7x-10-x^2}}+C[/tex] etc
Edit:Am gresit la substitutie in loc de x-7/2=9/4siny vine x-7/2=3/2siny.Asta nu schimba foarte mult ideea de rezolvare.

[juantheron]

thanks zec

[juantheron]

[tex]\displaystyle \int_{\frac{1}{2}}^{1} \frac{\sqrt{x-x^3}}{x^4}dx[/tex]