por liviabgomes » Qua Jun 01, 2011 15:14
![f(x)=\int_{0}^\(pi/2)_ \](/latexrender/pictures/6968c95674f870c730dd2337ca2698cc.png "f(x)=\int_{0}^\(pi/2)_ \")
sen(x) cos²(x) dx=
por carlosalesouza » Qua Jun 01, 2011 16:30
![\\
sen(x).cos^2(x)\\
\left [sen(x).cos(x)\right ].cos(x)\\
\left [ \frac{1}{2}sen(2x)\right].cos(x)\\
\frac{1}{2}[sen(2x).cos(x)]\\
\frac{1}{2}\left[\frac{1}{2}(sen(x+2x)-sen(x-2x))\right ]\\
\frac{1}{4}[sen(3x)-sen(-x)]\\
\frac{1}{4}[sen(3x)+sen(x)]](/latexrender/pictures/a3553111601b4164756c1f342ca143a5.png "\\
sen(x).cos^2(x)\\
\left [sen(x).cos(x)\right ].cos(x)\\
\left [ \frac{1}{2}sen(2x)\right].cos(x)\\
\frac{1}{2}[sen(2x).cos(x)]\\
\frac{1}{2}\left[\frac{1}{2}(sen(x+2x)-sen(x-2x))\right ]\\
\frac{1}{4}[sen(3x)-sen(-x)]\\
\frac{1}{4}[sen(3x)+sen(x)]")
![\\
\displaystyle \int_0^\frac{\pi}{2}[sen(3x)+sen(x)]dx](/latexrender/pictures/552ebeee3f2d9e58580e76e20ea4868f.png "\\
\displaystyle \int_0^\frac{\pi}{2}[sen(3x)+sen(x)]dx")
![\\
\left [-cos(3x)-cos(x)\right ]_0^\frac{\pi}{2}\\
\left [-cos\left(3.\frac{\pi}{2}\right )-cos\left(\frac{\pi}{2}\right)\right ]-[-cos(3.0)-cos(0)]\\](/latexrender/pictures/88d788c966f6b5ee276e788add593252.png "\\
\left [-cos(3x)-cos(x)\right ]_0^\frac{\pi}{2}\\
\left [-cos\left(3.\frac{\pi}{2}\right )-cos\left(\frac{\pi}{2}\right)\right ]-[-cos(3.0)-cos(0)]\\")
![\\
\left [-(0)-(0)\right ]-\left [-(1)-(1)\right ]\\
0-(-2) = 2](/latexrender/pictures/9ff3f3fe5b993d4ab4f3b3d746b34351.png "\\
\left [-(0)-(0)\right ]-\left [-(1)-(1)\right ]\\
0-(-2) = 2")
por LuizAquino » Qua Jun 01, 2011 16:36
.
, com a real não nulo.
por liviabgomes » Qua Jun 01, 2011 17:54
por liviabgomes » Qua Jun 01, 2011 17:58
por LuizAquino » Qua Jun 01, 2011 18:28
![\int_0^\frac{\pi}{2} \textrm{sen}\,x\cos^2 x \,dx = \left[- \frac{1}{3} \cos^3 x\right]_0^\frac{\pi}{2} = \left(- \frac{1}{3} \cos^3 \frac{\pi}{2}\right) - \left(- \frac{1}{3} \cos^3 0\right) = \frac{1}{3}](/latexrender/pictures/95a03fd0e7b6c22fbfc47108025bf29e.png "\int_0^\frac{\pi}{2} \textrm{sen}\,x\cos^2 x \,dx = \left[- \frac{1}{3} \cos^3 x\right]_0^\frac{\pi}{2} = \left(- \frac{1}{3} \cos^3 \frac{\pi}{2}\right) - \left(- \frac{1}{3} \cos^3 0\right) = \frac{1}{3}")
por carlosalesouza » Qui Jun 02, 2011 08:53
por LuizAquino » Qui Jun 02, 2011 12:16
![\int_0^\frac{\pi}{2} \textrm{sen}\,x\cos^2 x \,dx = \frac{1}{4}\int_0^\frac{\pi}{2}\textrm{sen}\,3x + \textrm{sen}\,x \,dx = \left[-\frac{1}{12}\cos 3x - \frac{1}{4}\cos x\right]_0^\frac{\pi}{2}](/latexrender/pictures/705fc07dac52aa016f39f075a31a2f24.png "\int_0^\frac{\pi}{2} \textrm{sen}\,x\cos^2 x \,dx = \frac{1}{4}\int_0^\frac{\pi}{2}\textrm{sen}\,3x + \textrm{sen}\,x \,dx = \left[-\frac{1}{12}\cos 3x - \frac{1}{4}\cos x\right]_0^\frac{\pi}{2}")
por carlosalesouza » Qui Jun 02, 2011 12:38
Voltar para Cálculo: Limites, Derivadas e Integrais
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![\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}](/latexrender/pictures/981987c7bcdf9f8f498ca4605785636a.png "\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}")
![\sqrt[]{2}](/latexrender/pictures/f21662d1cabab6e8b273a4b6f1cd663a.png "\sqrt[]{2}")
(dica : igualar a expressão a 
e elevar ao quadrado os dois lados)