$int_(0)^(15)x^2sqrt(a^2-x^2)dx$ ?

$int_0^a x^2sqrt(a^2-x^2)dx = (a^4pi)/16$

Avalie:

$int_0^a x^2sqrt(a^2-x^2)dx$

Substituto:

$x= asint$

$dx = a costdt$

com $t in [0,pi/2]$

de modo a:

$int_0^a x^2sqrt(a^2-x^2)dx = int_0^(pi/2) a^2 sin^2t sqrt(a^2-a^2 sin^2t)acostdt$

$int_0^a x^2sqrt(a^2-x^2)dx = a^4 int_0^(pi/2) sin^2t sqrt(1- sin^2t)costdt$

Para se qualificar para o $t in [0,pi/2]$ o cosseno é positivo, então:

$sqrt(1- sin^2t) = cost$

e depois:

$int_0^a x^2sqrt(a^2-x^2)dx = a^4 int_0^(pi/2) sin^2t cos^2tdt$

Use agora as identidades trigonométricas:

$sin 2theta = 2 sin theta cos theta$

$2sin^2 theta = 1- cos theta$

Sun:

$sin^2t cos^2tdt = 1/4 sin^2 2t = 1/8(1-cos4t)$

$int_0^a x^2sqrt(a^2-x^2)dx = a^4/8 int_0^(pi/2) (1-cos4t)dt$

Usando agora a linearidade da integral:

$int_0^a x^2sqrt(a^2-x^2)dx = a^4/8 (int_0^(pi/2) dt - int_0^(pi/2) cos4tdt)$

$int_0^a x^2sqrt(a^2-x^2)dx = a^4/8 [t - (sin 4t)/4]_0^(pi/2)$

$int_0^a x^2sqrt(a^2-x^2)dx = (a^4pi)/16$

int_(0)^(15)x^2sqrt(a^2-x^2)dxint_(0)^(15)x^2sqrt(a^2-x^2)dx?