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substituição trigonométrica ∫50x3√1−25x2dx
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Solution
225(−13(1−25x2)32+15(1−25x2)52)+C
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Solution steps
Solve by:
One step at a time
∫50x3√1−25x2dx
Remover a constante: ∫a·f(x)dx=a·∫f(x)dx
=50·∫x3√1−25x2dx
Aplicar integração por substituição trigonométrica:∫1625sin3(u)cos2(u)du
=50·∫1625sin3(u)cos2(u)du
Remover a constante: ∫a·f(x)dx=a·∫f(x)dx
=50·1625·∫sin3(u)cos2(u)du
Reeecreva usando identidades trigonométricas
=50·1625·∫(1−cos2(u))sin(u)cos2(u)du
Aplicar integração por substituição:∫−v2(1−v2)dv
=50·1625·∫−v2(1−v2)dv
Expandir −v2(1−v2):−v2+v4
=50·1625·∫−v2+v4dv
Aplicar a regra da soma: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx
