Question

Intregral Trigonométrica por substituição com produtos de potência de seno e cosseno:
∫sen^2x*cos^4xdx de 0 até pi?
me ajudem

Answer 1

Olá

Lembrando que: 

$sen^{2}x+cos^{2}x = 1$ 

$cos^{2}x = \frac{1+cos2x}{2}$

Temos que:

$\int\limits^\pi_0 {sen^{2}x.cos^{4}x} \, dx = \int\limits^\pi_0 {(1-cos^{2}x)cos^{4}x} \, dx = \int\limits^\pi_0 {cos^{4}x - cos^{6}x} \, dx =$

Daí temos que:

$= \int\limits^\pi_0 {( \frac{1+cos(2x)}{2})( \frac{1+cos(2x)}{2}) } \, dx - \int\limits^\pi_0 {( \frac{1+cos(2x)}{2})( \frac{1+cos(2x)}{2})( \frac{1+cos(2x)}{2})} \, dx$
$= \frac{1}{4} \int\limits^\pi_0 {(1+cos(2x))(1+cos(2x))} \, dx - \frac{1}{8} \int\limits^\pi_0 {(1+cos(2x))(1+cos(2x))(1+cos(2x))} \, dx$
$= \frac{1}{4} \int\limits^\pi_0 {1+2cos(2x)+cos^{2}(2x)} \, dx - \frac{1}{8} \int\limits^\pi_0 {cos^{3}(2x)+3cos^{2}(2x)+3cos(2x)+1} \, dx$

Resolvendo essas duas integrais, temos que:

$= \frac{3x}{8} + \frac{1}{4}sen(2x) + \frac{1}{32} sen(4x) - \frac{5x}{16} - \frac{15}{64}sen(2x) - \frac{3}{64}sen(4x)- \frac{1}{192}sen(6x)$

$= \frac{x}{16} + \frac{1}{64}sen(2x)- \frac{1}{64}sen(4x)- \frac{1}{192}sen(6x)$

Aplicando os limites temos que:

$= \frac{\pi}{6}$

Answer 2

Resposta:

Explicação passo-a-passo:

$\int _0^{\pi }\sin ^2\left(x\right)\cos ^4\left(x\right)dx$

Fazendo cos x = u

cos^4 u = u^4

sen² x = $\sqrt{1 - u^{2} }$

$=\int _1^{-1}-u^4\sqrt{1-u^2}du$

$=-\int _{-1}^1-u^4\sqrt{1-u^2}du$

Fazendo u = sen (v)

$=-\left(-\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin ^4\left(v\right)\cos ^2\left(v\right)dv\right)$

$=-\left(-\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin ^4\left(v\right)\left(1-\sin ^2\left(v\right)\right)dv\right)\\\\=-\left(-\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin ^4\left(v\right)-\sin ^6\left(v\right)dv\right)\\\\=-\left(-\left(\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin ^4\left(v\right)dv-\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin ^6\left(v\right)dv\right)\right)\\\\=-\left(-\left(\frac{3\pi }{8}-\frac{5\pi }{16}\right)\right)\\\\=\frac{\pi }{16}$