Métodos de integración: Sustitución trigonométrica - Integración producto de seno y coseno para resolverlos. RESOLVER CON SU PROCESO BIEN IDENTIFICADO LOS SIGUIENTE PROBLEMAS.
Nota: Ya añadi el documento en pdf. Y si no es mucha molestia enviar la solución y todo al correo: [email protected] por que mas tengo chance de revisar el correo por el telefono que por la computadora. Gracias de Antemano.
Adjuntos:
F4BI4N:
pide autenticación en la pag :v
Respuestas
Respuesta dada por:
1
por sustitucion trigonometrica
1)
![\int \frac{x^2}{\sqrt{25-x^2}}dx...................\:x=5\sin \left(u\right)......dx=5\cos \left(u\right)du \int \frac{x^2}{\sqrt{25-x^2}}dx...................\:x=5\sin \left(u\right)......dx=5\cos \left(u\right)du](https://tex.z-dn.net/?f=%5Cint+%5Cfrac%7Bx%5E2%7D%7B%5Csqrt%7B25-x%5E2%7D%7Ddx...................%5C%3Ax%3D5%5Csin+%5Cleft%28u%5Cright%29......dx%3D5%5Ccos+%5Cleft%28u%5Cright%29du)
reemplazando
![=\int \frac{\left(5\sin \left(u\right)\right)^2}{\sqrt{25-\left(5\sin \left(u\right)\right)^2}}5\cos \left(u\right)du =\int \frac{\left(5\sin \left(u\right)\right)^2}{\sqrt{25-\left(5\sin \left(u\right)\right)^2}}5\cos \left(u\right)du](https://tex.z-dn.net/?f=%3D%5Cint+%5Cfrac%7B%5Cleft%285%5Csin+%5Cleft%28u%5Cright%29%5Cright%29%5E2%7D%7B%5Csqrt%7B25-%5Cleft%285%5Csin+%5Cleft%28u%5Cright%29%5Cright%29%5E2%7D%7D5%5Ccos+%5Cleft%28u%5Cright%29du)
![=125\int \frac{\sin ^2\left(u\right)\cos \left(u\right)}{\sqrt{25-25\sin ^2\left(u\right)}}du=125\int \frac{\sin ^2\left(u\right)\cos \left(u\right)}{5\sqrt{-\sin ^2\left(u\right)+1}}du =125\int \frac{\sin ^2\left(u\right)\cos \left(u\right)}{\sqrt{25-25\sin ^2\left(u\right)}}du=125\int \frac{\sin ^2\left(u\right)\cos \left(u\right)}{5\sqrt{-\sin ^2\left(u\right)+1}}du](https://tex.z-dn.net/?f=%3D125%5Cint+%5Cfrac%7B%5Csin+%5E2%5Cleft%28u%5Cright%29%5Ccos+%5Cleft%28u%5Cright%29%7D%7B%5Csqrt%7B25-25%5Csin+%5E2%5Cleft%28u%5Cright%29%7D%7Ddu%3D125%5Cint+%5Cfrac%7B%5Csin+%5E2%5Cleft%28u%5Cright%29%5Ccos+%5Cleft%28u%5Cright%29%7D%7B5%5Csqrt%7B-%5Csin+%5E2%5Cleft%28u%5Cright%29%2B1%7D%7Ddu)
usamos lapitagorica para el denominador
![\:1-\sin ^2\left(x\right)=\cos ^2\left(x\right) \:1-\sin ^2\left(x\right)=\cos ^2\left(x\right)](https://tex.z-dn.net/?f=%5C%3A1-%5Csin+%5E2%5Cleft%28x%5Cright%29%3D%5Ccos+%5E2%5Cleft%28x%5Cright%29)
![= \frac{125}{5} \int \frac{\cos \left(u\right)\sin ^2\left(u\right)}{\sqrt{\cos ^2\left(u\right)}}du=25\int \sin ^2\left(u\right)du = \frac{125}{5} \int \frac{\cos \left(u\right)\sin ^2\left(u\right)}{\sqrt{\cos ^2\left(u\right)}}du=25\int \sin ^2\left(u\right)du](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B125%7D%7B5%7D+%5Cint+%5Cfrac%7B%5Ccos+%5Cleft%28u%5Cright%29%5Csin+%5E2%5Cleft%28u%5Cright%29%7D%7B%5Csqrt%7B%5Ccos+%5E2%5Cleft%28u%5Cright%29%7D%7Ddu%3D25%5Cint+%5Csin+%5E2%5Cleft%28u%5Cright%29du)
con la identidad
![\sin ^2\left(x\right)=\frac{1-\cos \left(2x\right)}{2} \sin ^2\left(x\right)=\frac{1-\cos \left(2x\right)}{2}](https://tex.z-dn.net/?f=%5Csin+%5E2%5Cleft%28x%5Cright%29%3D%5Cfrac%7B1-%5Ccos+%5Cleft%282x%5Cright%29%7D%7B2%7D)
![=25\int \frac{1-\cos \left(2u\right)}{2}du= \frac{25}{2}(\int \:1du-\int \cos \left(2u\right)du)=\frac{25}{2}\left(u-\frac{1}{2}\sin \left(2u\right)\right) =25\int \frac{1-\cos \left(2u\right)}{2}du= \frac{25}{2}(\int \:1du-\int \cos \left(2u\right)du)=\frac{25}{2}\left(u-\frac{1}{2}\sin \left(2u\right)\right)](https://tex.z-dn.net/?f=%3D25%5Cint+%5Cfrac%7B1-%5Ccos+%5Cleft%282u%5Cright%29%7D%7B2%7Ddu%3D+%5Cfrac%7B25%7D%7B2%7D%28%5Cint+%5C%3A1du-%5Cint+%5Ccos+%5Cleft%282u%5Cright%29du%29%3D%5Cfrac%7B25%7D%7B2%7D%5Cleft%28u-%5Cfrac%7B1%7D%7B2%7D%5Csin+%5Cleft%282u%5Cright%29%5Cright%29+)
![x=5sen^2(u).............u=\arcsin \left(\frac{1}{5}x\right) x=5sen^2(u).............u=\arcsin \left(\frac{1}{5}x\right)](https://tex.z-dn.net/?f=x%3D5sen%5E2%28u%29.............u%3D%5Carcsin+%5Cleft%28%5Cfrac%7B1%7D%7B5%7Dx%5Cright%29)
![=\frac{25}{2}\left(\arcsin \left(\frac{x}{5}\right)-\frac{1}{2}\sin \left(2\arcsin \left(\frac{x}{5}\right)\right)\right)+k =\frac{25}{2}\left(\arcsin \left(\frac{x}{5}\right)-\frac{1}{2}\sin \left(2\arcsin \left(\frac{x}{5}\right)\right)\right)+k](https://tex.z-dn.net/?f=%3D%5Cfrac%7B25%7D%7B2%7D%5Cleft%28%5Carcsin+%5Cleft%28%5Cfrac%7Bx%7D%7B5%7D%5Cright%29-%5Cfrac%7B1%7D%7B2%7D%5Csin+%5Cleft%282%5Carcsin+%5Cleft%28%5Cfrac%7Bx%7D%7B5%7D%5Cright%29%5Cright%29%5Cright%29%2Bk)
2)
![\int \frac{1}{\left(1+x^2\right)^{\frac{3}{2}}}dx...........x=tan(u)....\:dx=\sec ^2\left(u\right)du \int \frac{1}{\left(1+x^2\right)^{\frac{3}{2}}}dx...........x=tan(u)....\:dx=\sec ^2\left(u\right)du](https://tex.z-dn.net/?f=%5Cint+%5Cfrac%7B1%7D%7B%5Cleft%281%2Bx%5E2%5Cright%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7Ddx...........x%3Dtan%28u%29....%5C%3Adx%3D%5Csec+%5E2%5Cleft%28u%5Cright%29du)
![=\int \frac{1}{\left(1+\tan ^2\left(u\right)\right)^{\frac{3}{2}}}\sec ^2\left(u\right)du=\int \frac{\sec ^2\left(u\right)}{\left(\sec ^2\left(u\right)\right)^{\frac{3}{2}}}du =\int \frac{1}{\left(1+\tan ^2\left(u\right)\right)^{\frac{3}{2}}}\sec ^2\left(u\right)du=\int \frac{\sec ^2\left(u\right)}{\left(\sec ^2\left(u\right)\right)^{\frac{3}{2}}}du](https://tex.z-dn.net/?f=%3D%5Cint+%5Cfrac%7B1%7D%7B%5Cleft%281%2B%5Ctan+%5E2%5Cleft%28u%5Cright%29%5Cright%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%5Csec+%5E2%5Cleft%28u%5Cright%29du%3D%5Cint+%5Cfrac%7B%5Csec+%5E2%5Cleft%28u%5Cright%29%7D%7B%5Cleft%28%5Csec+%5E2%5Cleft%28u%5Cright%29%5Cright%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7Ddu)
![=\int \frac{\sec ^2\left(u\right)}{\sec ^3\left(u\right)}du=\int \frac{1}{\sec \left(u\right)}du=\int \cos \left(u\right)du=sen(u)
=\int \frac{\sec ^2\left(u\right)}{\sec ^3\left(u\right)}du=\int \frac{1}{\sec \left(u\right)}du=\int \cos \left(u\right)du=sen(u)](https://tex.z-dn.net/?f=%3D%5Cint+%5Cfrac%7B%5Csec+%5E2%5Cleft%28u%5Cright%29%7D%7B%5Csec+%5E3%5Cleft%28u%5Cright%29%7Ddu%3D%5Cint+%5Cfrac%7B1%7D%7B%5Csec+%5Cleft%28u%5Cright%29%7Ddu%3D%5Cint+%5Ccos+%5Cleft%28u%5Cright%29du%3Dsen%28u%29%0A)
como
![\:u=\arctan \left(x\right).. \:u=\arctan \left(x\right)..](https://tex.z-dn.net/?f=%5C%3Au%3D%5Carctan+%5Cleft%28x%5Cright%29..)
![=\sin \left(\arctan \left(x\right)\right)=\frac{x}{\sqrt{x^2+1}}+k =\sin \left(\arctan \left(x\right)\right)=\frac{x}{\sqrt{x^2+1}}+k](https://tex.z-dn.net/?f=%3D%5Csin+%5Cleft%28%5Carctan+%5Cleft%28x%5Cright%29%5Cright%29%3D%5Cfrac%7Bx%7D%7B%5Csqrt%7Bx%5E2%2B1%7D%7D%2Bk++)
3)![\int \frac{\sqrt{x^2-9}}{x^4}dx.........x=3\sec \left(u\right) ,,,,dx=3sec(u)tan(u) \int \frac{\sqrt{x^2-9}}{x^4}dx.........x=3\sec \left(u\right) ,,,,dx=3sec(u)tan(u)](https://tex.z-dn.net/?f=%5Cint+%5Cfrac%7B%5Csqrt%7Bx%5E2-9%7D%7D%7Bx%5E4%7Ddx.........x%3D3%5Csec+%5Cleft%28u%5Cright%29++%2C%2C%2C%2Cdx%3D3sec%28u%29tan%28u%29)
4)![\int \frac{\sqrt{9-x^2}}{x}dx...........\:u=9-x^2......... du=-2xdx \int \frac{\sqrt{9-x^2}}{x}dx...........\:u=9-x^2......... du=-2xdx](https://tex.z-dn.net/?f=%5Cint+%5Cfrac%7B%5Csqrt%7B9-x%5E2%7D%7D%7Bx%7Ddx...........%5C%3Au%3D9-x%5E2.........+++++du%3D-2xdx)
![\int \frac{\sqrt{u}}{-2}du=\int \frac{u^{\frac{1}{2}}}{-2}du \int \frac{\sqrt{u}}{-2}du=\int \frac{u^{\frac{1}{2}}}{-2}du](https://tex.z-dn.net/?f=%5Cint+%5Cfrac%7B%5Csqrt%7Bu%7D%7D%7B-2%7Ddu%3D%5Cint+%5Cfrac%7Bu%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B-2%7Ddu)
![=\frac{1}{-2}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}=-\frac{u^{\frac{3}{2}}}{3}+C =\frac{1}{-2}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}=-\frac{u^{\frac{3}{2}}}{3}+C](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B-2%7D%5Cfrac%7Bu%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B%5Cfrac%7B3%7D%7B2%7D%7D%3D-%5Cfrac%7Bu%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B3%7D%2BC)
![-\frac{\left(9-x^2\right)^{\frac{3}{2}}}{3}+C -\frac{\left(9-x^2\right)^{\frac{3}{2}}}{3}+C](https://tex.z-dn.net/?f=-%5Cfrac%7B%5Cleft%289-x%5E2%5Cright%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B3%7D%2BC)
1)
reemplazando
usamos lapitagorica para el denominador
con la identidad
2)
como
3)
4)
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