Sr Examen
Integral de b*cos(t)^2*(-3b)*sin(t)^4 dt
Solución
Ha introducido
[src]
0 / | | 2 4 | b*cos (t)*-3*b*sin (t) dt | / pi -- 2
$$\int\limits_{\frac{\pi}{2}}^{0} - 3 b b \cos^{2}{\left(t \right)} \sin^{4}{\left(t \right)}\, dt$$
Integral(((b*cos(t)^2)*(-3*b))*sin(t)^4, (t, pi/2, 0))
Respuesta (Indefinida)
[src]
/ | / 3 3 5 6 6 5 2 4 4 2 \ | 2 4 2 | cos (t)*sin (t) cos (t)*sin(t) t*cos (t) t*sin (t) sin (t)*cos(t) 3*t*cos (t)*sin (t) 3*t*cos (t)*sin (t)| | b*cos (t)*-3*b*sin (t) dt = C - 3*b *|- --------------- - -------------- + --------- + --------- + -------------- + ------------------- + -------------------| | \ 6 16 16 16 16 16 16 / /
$$\int - 3 b b \cos^{2}{\left(t \right)} \sin^{4}{\left(t \right)}\, dt = C - 3 b^{2} \left(\frac{t \sin^{6}{\left(t \right)}}{16} + \frac{3 t \sin^{4}{\left(t \right)} \cos^{2}{\left(t \right)}}{16} + \frac{3 t \sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)}}{16} + \frac{t \cos^{6}{\left(t \right)}}{16} + \frac{\sin^{5}{\left(t \right)} \cos{\left(t \right)}}{16} - \frac{\sin^{3}{\left(t \right)} \cos^{3}{\left(t \right)}}{6} - \frac{\sin{\left(t \right)} \cos^{5}{\left(t \right)}}{16}\right)$$
Respuesta
[src]
2 3*pi*b ------- 32
$$\frac{3 \pi b^{2}}{32}$$
=
=
2 3*pi*b ------- 32
$$\frac{3 \pi b^{2}}{32}$$
3*pi*b^2/32
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.