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

Limits, differentiation, related rates, integration, trig integrals, etc.
davidherrera
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Joined: Mon Apr 21, 2014 11:28 pm
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### Is there another way to solve this integral????

Hi, everyone I'm a new guy in these forums my name is David . 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!!

$\int_{}^{} sen^2(2x)cos^2(2x)\, dx$
Using the following identitites \ Usando las siguientes identidades
$sen^2(u) = \frac{1}{2}-\frac{1}{2}cos(2u)$
$cos^2(u) = \frac{1}{2}+\frac{1}{2}cos(2u)$
We write the new values in the original integral \ Sustituimos valores
$\int_{}^{} ( \frac{1}{2}-\frac{1}{2}cos(4x)) ( \frac{1}{2}+\frac{1}{2}cos(4x)) \, dx$
This is a difference of squares \ Esto es una diferencia de cuadrados
$( a + b )( a - b ) = a^2 + b^2$
Reducing \ Reduciendo
$\int_{}^{} \frac{1}{4}- \frac{1}{4}cos^2(4x) \, dx$
We observe that we have again a cosine with a potency $cos^2(4x)$, and we can´t solve our integral with this expression inside it.
Observamos que otra vez tenemos un coseno con potencia $cos^2(4x)$, 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
$cos^2(4x) = \frac{1}{2}+ \frac{1}{2}cos(8x)$
putting this value on my integral \ poniendo esto en mi integral
$\int_{}^{} \frac{1}{4}- \frac{1}{4} ( \frac{1}{2}+ \frac{1}{2}cos(8x) ) \, dx$
Reducing \ Reduciendo
$\int_{}^{} \frac{1}{4}- \frac{1}{8} - \frac{1}{8}cos(8x) \, dx == \int_{}^{} \frac{1}{8} - \frac{1}{8}cos(8x) \, dx$
And we can divide in two integrals \ Y ahora podemos dividir en dos integrales
$\frac{1}{8}\int_{}^{} \, dx - \frac{1}{8} \int_{}^{} cos(8x) \, dx$
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
$\frac{1}{8}\int_{}^{} \, dx - \frac{1}{8} \int_{}^{} cos(8x) \, dx$
$u= 8x$
$du= 8 dx$
$dx =\frac{du}{8}$
Replacing values \ Reemplazando valores
$\frac{1}{8}\int_{}^{} \, dx - \frac{1}{8} \int_{}^{} cos(u) \frac{du}{8} \,$
Taking out the constant of the second integral \ Sacando la constante de la segunda integral
$\frac{1}{8}\int_{}^{} \, dx - \frac{1}{8} . \frac{1}{8}\int_{}^{} cos(u) \,du$
We have \ Tenemos
$\frac{1}{8}\int_{}^{} \, dx - \frac{1}{16}\int_{}^{} cos(u) \,du$
Finally we solve this two easy integrals and replace the value of u / Finalmente resolvemos y remplazamos el valor de u
$\frac{1}{8} x - \frac{1}{16} sen(u)$
$\frac{1}{8} x - \frac{1}{16} sen(8x)$
Finally our answer is!!! \ Finalmente nuestra respuesta es!!!
$\int_{}^{} sen^2(2x)cos^2(2x)\, dx == \frac{1}{8} x - \frac{1}{16} sen(8x) + C$

buddy
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Joined: Sun Feb 22, 2009 10:05 pm
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### Re: Is there another way to solve this integral????

davidherrera wrote: $\int_{}^{} sen^2(2x)cos^2(2x)\, dx$

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