Is there another way to solve this integral????

Limits, differentiation, related rates, integration, trig integrals, etc.

Is there another way to solve this integral????

Postby davidherrera on Tue Apr 22, 2014 2:10 am

Hi, everyone I'm a new guy in these forums my name is David :wave: . I am a student of the National Autonomous University of Mexico and I live in Mexico City. Our calculus teacher Pablo left us a special job. We have to ask people from other countries if we have another way to solve some integrals....I am skeptical about internet forums...but I'm looking for answers in all media... however... I do not lose anything by trying...
I´ll let my integral with my own solution, but if you find another way to get the same result please let your procedure here...I'll be very grateful...
IF YOU CAN TELL ME WHICH COUNTRY ARE YOU FROM THAT WOULD WE AWESOME!! :clap:


Using the following identitites \ Usando las siguientes identidades


We write the new values in the original integral \ Sustituimos valores

This is a difference of squares \ Esto es una diferencia de cuadrados

Reducing \ Reduciendo

We observe that we have again a cosine with a potency , and we can´t solve our integral with this expression inside it.
Observamos que otra vez tenemos un coseno con potencia , y no podemos resolver nuestra integral con esta expresión dentro de ella.
so... using the previus properties we know that \ así que... usando las propiedades anteriores sabemos que

putting this value on my integral \ poniendo esto en mi integral

Reducing \ Reduciendo

And we can divide in two integrals \ Y ahora podemos dividir en dos integrales

The first of ther is very easy, but for the second one we must have to do a change of variable
La primera es muy fácil, pero para la segunda debemos hacer cambio de variable




Replacing values \ Reemplazando valores

Taking out the constant of the second integral \ Sacando la constante de la segunda integral

We have \ Tenemos

Finally we solve this two easy integrals and replace the value of u / Finalmente resolvemos y remplazamos el valor de u


Finally our answer is!!! \ Finalmente nuestra respuesta es!!!
davidherrera
 
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Re: Is there another way to solve this integral????

Postby buddy on Tue Apr 22, 2014 12:21 pm

davidherrera wrote:

you could also do like:

sin^2(2x)cos^2(2x)
=[sin(2x)cos(2x)]^2
=[(1/2)(2/1)sin(2x)cos(2x)]^2
=[1/2]^2 [2sin(2x)cos(2x)]^2
=(1/4)[sin(4x)]^2
=(1/4)sin^2(4x)

cos(2x)=1-2sin^2(x)
2sin^2(x)=1-cos(2x)
sin^2(x)=(1/2)(1-cos(2x))

so

(1/4)sin^2(4x)
=(1/4)(1/2)(1-cos(8x))
=(1/8)(1-cos(8x))
=1/8 - cos(8x)/8

then integrate
buddy
 
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