gostaria de entender o pq de (sen(PI/2)+senx)=2sen(PI/4+x/2).COS(PI/4-x/2)como vc chegou nisso? obg
Olá, Leandro!Sejam A e B dois ângulos quaisquer. Então:sen(A+B) + sen(A-B) == (senA.cosB + senB.cosA) + (senA.cosB - senB.cosA) == 2.senA.cosBFaçamos as seguintes transformações:x = A+By = A-BSomando as duas equações:x+y = 2A --> A = (x+y)/2Subtraindo as duas equações:x-y = 2B --> B = (x-y)/2Assim, substituindo A e B em função de x e y na identidade senA + senB = 2.senA.cosB, temos:sen(x) + sen(y) = 2.sen[(x+y)/2].cos[(x-y)/2]Substituindo os valores de x e y chegamos ao resultado.Agradecemos pelo comentário e volte sempre!
oi, obg por responder, gostaria de saber se a minha solução para o problema também é aceitável aqui vai: ∫ dx/(1+senx)²= ∫ [(1-senx)(1-senx)]/[(1+senx)(1+senx)(1-senx)(1-senx)]dx(multipliquei encima e embaixo por (1-senx)²)=∫(1-senx)²/(1-sen²x)²dx=∫(1-senx)²/(cos²x)²dx=∫(1-senx)²/(cosx)^4dx=∫(1-2senx+sen²x)/(cosx)^4dx=∫ 1/(cosx)^4dx -2∫senx/(cosx)^4dx +∫(sen²x)/(cosx)^4dx∫(secx)^4dx -2∫senx/(cosx)^4dx +∫(sen²x)/(cosx)^4dxchamando :∫(secx)^4dx=(I)∫senx/(cosx)^4dx =(II)∫(sen²x)/(cosx)^4dx=(III)resolvendo (I)∫(secx)^4dx=∫(sec²x.sec²x)dxfazendo u=sec²x => du=2sec²x.tgxdx, v=sec²xdx,dv=tgxentão:∫(secx)^4dx=(secx)^4-2∫(secx)^4.tgxdx∫(secx)^4.tgxdx=∫(secx)².(secx)².tgxdx=∫(secx)²(1+tg²x).tgxdxchamando tgx=u =>du=sec²xdx∫(1+u²)udu=∫(u+u³)du=u²/2 +u^4/4=tg²x/2 +tg^4x/4 +clogo : ∫(secx)^4.tgxdx=tg²x/2 +tg^4x/4 +c, então∫(secx)^4dx=(secx)^4-2∫(secx)^4.tgxdx=(secx)^4-2(tg²x/2 +tg^4x/4)=(secx)^4-tg²x +tg^4x/2+cresolvendo (II)∫senx/(cosx)^4dx =chamando cosx=u => du=-senxdx∫senx/(cosx)^4dx=∫-du/u^4=-∫u^-4du=-u^(-3)/-3=1/3u³=1/3(cosx)³ +cresolvendo (III)∫(sen²x)/(cosx)^4dx=∫tg²x.sec²xdx=chamando tgx=u => du=sec²xdx∫tg²x.sec²xdx=∫u²du=u³/3=tg³x/3+clogo∫(secx)^4dx -2∫senx/(cosx)^4dx +∫(sen²x)/(cosx)^4dx(secx)^4-tg²x +tg^4x/2-2/3(cosx)³+tg³x/3 +c da uma olhada ai :)
poe no papel que fica mais fácil de entender
![[;\int \frac{dx}{(1+sen(x))^2};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tvqEauYbkILsDKbQMxHkzkKvWMVTSmUafhtElHwFD31LYrfUNSjpjVFPEkFzCe8VOuGRY1dLvJ6_MMW_RoXQ-W2OntvuwWk1XdsUQUmSs5_Vxc14owEtOumRHeAFnZ7Dd-RNrs3tiuEdqaNNk=s0-d "\int \frac{dx}{(1+sen(x))^2}")
![[;\int \frac{dx}{(sen(\frac{\pi}{2})+sen(x))^2};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vX9Ux3sD1F_s080DPqBiuHrad2fOYtmKQfM5IT98Z0NQU_wS0QhlGGiNTHOD2MjN2d67YeQENo0VW6eYE8iHZ-hVt7Pti8ngLCH7cQUDlESrT-1CllAFGa-bqLUUOA_DGKgPnzs4wh5BKpiqal4wGpYnmea_nidJNkQEVjvKEcBnF6IWEyFKDKQCwSAso=s0-d "\int \frac{dx}{(sen(\frac{\pi}{2})+sen(x))^2}")
![[;\int \frac{dx}{(2sen(\frac{\pi}{4}+\frac{x}{2})cos(\frac{\pi}{4}-\frac{x}{2}))^2};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vDkLdhiA59gUMkteXFnAyOgH5zeIddWsb4vDDKZSiFOtPvznC1w--NxmroPPjlNJcRhVaGSxpKt2IKn9Cjrjtel3fjHhTNkUoTk8qNXvKDZs4GukTdNWwGRRHMf2uUi3Uej-NlYvohIAbgMJ5A7dL0RgvAvah6eo9yxSuoWckbVdfEB3tEX7dNiFCHEzNrRAZ1fAqe_mDuUWL-S_N3HvG_3EZSp_DJvA6MRaMgPH89Ax9Df_lWWwCXjAjnKqmYbPin_nwPIXHA7jGV4u5FCQntAQg=s0-d "\int \frac{dx}{(2sen(\frac{\pi}{4}+\frac{x}{2})cos(\frac{\pi}{4}-\frac{x}{2}))^2}")
![[;x=2u+\pi;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t6vjJfCi_FAl-LhyjK-KpPy3-aqN0QyI4z_6cEW1o6eRZX18KP7zd8IAE9TcxMral8XNmDP3ZzEsbw0C0k4yEjGWM=s0-d "x=2u+\pi")
![[;dx=2du;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sHTV2l6WbYCRnt5KkWK5jg_acFzJZ-kMwFCw9wQszkmTnGFl8jtWbXyjsV0rYNFP_wM51ywDjl2d7NM7xNMA=s0-d "dx=2du")
![[;\frac{1}{2}\int \frac{du}{(sen(\frac{\pi}{2}+\frac{\pi}{4}+u)cos(-\frac{\pi}{4}-u))^2}du;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vtQwZWGegnjxn2IHqYXC8v44GRIdWht3V9I4HcYPLjyNVCLIYdEXbr3iMUwofkceC1GVHuTOqDY_xZllOV6tXIHaSE6PU5MCDypbqXnSQ4MuqjH4ElBTzZ9RVlTTgSoncyGTDfE5H3R3g9SUi0MOCrbUpCtSzTqKgNI1V5YcAXhMyE-72Sv7gimt7ReULbuI07x3pqIan9RB0UcTAYL2HEVqXt9oS7RiR87-0z3p1rCswEu00Gf87IEjTkwwOCeEQbaz6lVgHjiX2md77BfYPLGk_aV5iDE9N-LUc=s0-d "\frac{1}{2}\int \frac{du}{(sen(\frac{\pi}{2}+\frac{\pi}{4}+u)cos(-\frac{\pi}{4}-u))^2}du")
![[;sen(\frac{\pi}{2}+\alpha)=cos(\alpha);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v1HjHOLJsX7fQRGamcA3gH7kLxbwaKtAXn6NpFy5JvniCYZZFQgj0q1_4UR8x0KKUlg1qgpEnJUozbeIOEkuVdg7H6ixz4pFW3iSYGwqNSGJHomVgYUK0fiZAN9g5YaqMcPTYcgfolmZaPeXEFeFl6lYyp-uc=s0-d "sen(\frac{\pi}{2}+\alpha)=cos(\alpha)")
![[;cos(-\alpha)=cos(\alpha);]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sbHi86w45GnuDDoNPe3e_131efSn5--YYKfjEAgY6WrwdOWcSOsjPYf9qFH3KlDlZbwIU7SrI855EJ1NrE1YtUKR3z47v82AtpmQXyxlEBsJoWJSSSECl4Qxor_A=s0-d "cos(-\alpha)=cos(\alpha)")
![[;\frac{1}{2}\int \frac{du}{(cos(\frac{\pi}{4}+u)cos(\frac{\pi}{4}+u))^2};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sI4Wnp9ZgiZ909vZu4Ra1cGsDRyOHG0SdLu8tGrjpq-fwOcqe-6_gz6zEMEuAD5W2z4t5-_G6ocGoMiYXcDx1cG5JZroCjvwNAoGgWUS_YwPQaeN-_Ft1QsHA-6SjLKdf6SzGCR0LrV5ofjPKYtU3e_J23VHgjGhX-dZqpppjpjqi6y6b8tRtqYZwyGzkT8vqE8mQ1wVcnrQtNHbsKJUL2l4e5TFK2PJZ7S8ZXBlEevxOGz2mHkd78Cb8OHe0=s0-d "\frac{1}{2}\int \frac{du}{(cos(\frac{\pi}{4}+u)cos(\frac{\pi}{4}+u))^2}")
![[;\frac{1}{2}\int sec^4(\frac{\pi}{4}+u)du;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v1pqUhe8ozKy5ZjnNfz_78bSU6yuaI77P5z35gdi_VypybrU0TdIElopQTN8nDqjS4dbqvBflhD58hWeyZCAI4FCw5bXJbozlVmx-nDSPbV3ccAnLjBWb4s926daSRpTLhhRhzROb8TuZyJGqZS3p4gR667azbtlqEzeFTVpJT=s0-d "\frac{1}{2}\int sec^4(\frac{\pi}{4}+u)du")
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:![[;\frac{1}{2}\int sec^4(z)dz;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_umYOgKJTaj03m1cC3SJ_ifz9QbstzUfCfoK6TLHh3g4YDSW_vv3DLgVJT3EzZaykBuZMFgiGZZkZC4E3gZCqjAEJmP-UNOTwlG-QDoSQ9Uffkt5FSz1iyYcxGFld3pTDNC4xnZWA=s0-d "\frac{1}{2}\int sec^4(z)dz")
![[;\frac{1}{6}\tan(z)(sec^2(z)+2)+C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uhC_iW4agOOHnnrstkO8NMLjFj9ExkchvMaCQJcq6V0zI8SwE1ry7vorcM-so1dHkaFAp2pEPy6wVdj2EBc9bjRHlK58Ec3u4E4ezB3MimsEDe-IIJu3Hcyk74kwkbRVRcLbjNrFiDCq7BZvtMT-wZiA=s0-d "\frac{1}{6}\tan(z)(sec^2(z)+2)+C")
![[;\int \frac{dx}{(1+sen(x))^2}= \frac{1}{6}\tan(\frac{x}{2}-\frac{\pi}{4})(sec^2(\frac{x}{2}-\frac{\pi}{4})+2)+C;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u6Nk2z59YVdMa54pMBXv1kgu9tKjTdh33OdMoiAzsclAfxA7zBARh9SCTBthMzdLgRApr7k2Se5A_SoP9pC89uLicEB71ev33Yrwt5KZJgIUCFQf887aXwyU4yGo4q9hvzskCFRXxHCDgV55HcD3pmfq-eviNyWyeeoac05lrL9Mxv1QvSSNZfqYkqd8ckri7l8z3Eyc-qMCStASjAuuoAjeEOXh0nrSWw1op5V2-gbyrlwdnEgIXokf2NYGkdD9r71MVSQ-62GswQOkKTabfR7uQyrjTTH1xMBm8g320WZe8FOyUNTyR3wujBMvs-MiIBcDr2ZUW0ZMfuZHu5odHjUn0fcrT-Dw=s0-d "\int \frac{dx}{(1+sen(x))^2}= \frac{1}{6}\tan(\frac{x}{2}-\frac{\pi}{4})(sec^2(\frac{x}{2}-\frac{\pi}{4})+2)+C")
gostaria de entender o pq de (sen(PI/2)+senx)=2sen(PI/4+x/2).COS(PI/4-x/2)
ResponderExcluircomo vc chegou nisso? obg
Olá, Leandro!
ResponderExcluirSejam A e B dois ângulos quaisquer. Então:
sen(A+B) + sen(A-B) =
= (senA.cosB + senB.cosA) + (senA.cosB - senB.cosA) =
= 2.senA.cosB
Façamos as seguintes transformações:
x = A+B
y = A-B
Somando as duas equações:
x+y = 2A --> A = (x+y)/2
Subtraindo as duas equações:
x-y = 2B --> B = (x-y)/2
Assim, substituindo A e B em função de x e y na identidade senA + senB = 2.senA.cosB, temos:
sen(x) + sen(y) = 2.sen[(x+y)/2].cos[(x-y)/2]
Substituindo os valores de x e y chegamos ao resultado.
Agradecemos pelo comentário e volte sempre!
oi, obg por responder, gostaria de saber se a minha solução para o problema também é aceitável aqui vai:
Excluir∫ dx/(1+senx)²=
∫ [(1-senx)(1-senx)]/[(1+senx)(1+senx)(1-senx)(1-senx)]dx(multipliquei encima e embaixo por (1-senx)²)=
∫(1-senx)²/(1-sen²x)²dx=
∫(1-senx)²/(cos²x)²dx=
∫(1-senx)²/(cosx)^4dx=
∫(1-2senx+sen²x)/(cosx)^4dx=
∫ 1/(cosx)^4dx -2∫senx/(cosx)^4dx +∫(sen²x)/(cosx)^4dx
∫(secx)^4dx -2∫senx/(cosx)^4dx +∫(sen²x)/(cosx)^4dx
chamando :
∫(secx)^4dx=(I)
∫senx/(cosx)^4dx =(II)
∫(sen²x)/(cosx)^4dx=(III)
resolvendo (I)
∫(secx)^4dx=∫(sec²x.sec²x)dx
fazendo u=sec²x => du=2sec²x.tgxdx, v=sec²xdx,dv=tgx
então:
∫(secx)^4dx=(secx)^4-2∫(secx)^4.tgxdx
∫(secx)^4.tgxdx=∫(secx)².(secx)².tgxdx=
∫(secx)²(1+tg²x).tgxdx
chamando tgx=u =>du=sec²xdx
∫(1+u²)udu=∫(u+u³)du=u²/2 +u^4/4=tg²x/2 +tg^4x/4 +c
logo : ∫(secx)^4.tgxdx=tg²x/2 +tg^4x/4 +c, então
∫(secx)^4dx=(secx)^4-2∫(secx)^4.tgxdx=
(secx)^4-2(tg²x/2 +tg^4x/4)=
(secx)^4-tg²x +tg^4x/2+c
resolvendo (II)
∫senx/(cosx)^4dx =
chamando cosx=u => du=-senxdx
∫senx/(cosx)^4dx=∫-du/u^4=-∫u^-4du=-u^(-3)/-3=1/3u³=1/3(cosx)³ +c
resolvendo (III)
∫(sen²x)/(cosx)^4dx=∫tg²x.sec²xdx=
chamando tgx=u => du=sec²xdx
∫tg²x.sec²xdx=∫u²du=u³/3=tg³x/3+c
logo
∫(secx)^4dx -2∫senx/(cosx)^4dx +∫(sen²x)/(cosx)^4dx
(secx)^4-tg²x +tg^4x/2-2/3(cosx)³+tg³x/3 +c
da uma olhada ai :)
poe no papel que fica mais fácil de entender
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