anonymous
  • anonymous
integrate 7x cos^4(x^2) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Try \[ u=x^2\quad du=2x\;dx \]
anonymous
  • anonymous
Also \[ \cos^2(u) = \frac{1+\cos(2u)}{2} \]
anonymous
  • anonymous
The only question is, how badly do you want it?

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anonymous
  • anonymous
The 7x is confusing me
anonymous
  • anonymous
My instinct says to solve it like an integration by parts
anonymous
  • anonymous
Okay, here is some basic algebra: \(7x=3.5\cdot 2x\)
anonymous
  • anonymous
\[ 7x = \frac 7 2 \cdot 2x \]
anonymous
  • anonymous
\[ \int 7x \cos^4(x^2) dx = \int \frac 7 2 \cos^4(x^2)\;2xdx \]
anonymous
  • anonymous
It's not like there is any obvious pattern here or anything.\[ u = x^2\quad du=2xdx \]
anonymous
  • anonymous
ah, okay
anonymous
  • anonymous
\[\int\limits_{}^{}\frac{ 7 }{ 2 }\cos ^{4}(u) du\]
anonymous
  • anonymous
?
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @wio Also \[ \cos^2(u) = \frac{1+\cos(2u)}{2} \] \(\color{blue}{\text{End of Quote}}\)
anonymous
  • anonymous
so you plug that in for cos and square the whole formula?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
and bring out the 7/2 constant
anonymous
  • anonymous
Sure.
anonymous
  • anonymous
\[\frac{ 7 }{ 2 }\int\limits_{}^{}(\frac{ 1+\cos(2u) }{ 1 })^{2}u du\]
anonymous
  • anonymous
Not \(udu\), just \(du\)
anonymous
  • anonymous
2 in the denominator
anonymous
  • anonymous
Now use foil.
anonymous
  • anonymous
what happened to the x^2 in cos^4x^2?
anonymous
  • anonymous
\(u=x^2\)
anonymous
  • anonymous
I'm very confused for some reason. Is the half angle formula squared because of the x^2 or the cos^4?
anonymous
  • anonymous
I was thinking that it was squared because of the 4th root on cos. That's why I put u du on the end and not just du
anonymous
  • anonymous
It's because the \(\cos\) is squared.
anonymous
  • anonymous
oh, I think it just clicked
anonymous
  • anonymous
okay
anonymous
  • anonymous
\[ \cos^4(u) = [\cos^2(u)]^2 = \left[\frac{1+\cos(2u)}{2}\right]^2 \]
anonymous
  • anonymous
so now we're multiplying it out
anonymous
  • anonymous
\[\frac{ 7 }{ 2 }\int\limits_{}^{}\frac{ 1+2\cos(2u)+\cos ^{2(2u)} }{ 4} du\]
anonymous
  • anonymous
not an exponent on the "2u" obviously
anonymous
  • anonymous
Pull out the 4 and use the half angle again on the squared cosine. Then it will be simple.
anonymous
  • anonymous
okay, I got this
anonymous
  • anonymous
I doubt I did this right\[\frac{ 14 }{ 16 }\int\limits_{}^{}2+2\cos(2u) du\]
anonymous
  • anonymous
Show your work.
anonymous
  • anonymous
\[\frac{ 7 }{ 8 }\int\limits_{}^{} 1+2\cos(2u)+(\frac{ 1+\cos(2u) }{ 2 }) du\]
anonymous
  • anonymous
should be \(4u\)
anonymous
  • anonymous
on the first cos?
anonymous
  • anonymous
Okay, which one do you think I was talking about?
anonymous
  • anonymous
the first one. I didn't have to do anything with the second besides plug in the formula
anonymous
  • anonymous
nevermind
anonymous
  • anonymous
I dont see why either one would be 4u
anonymous
  • anonymous
and I wouldn't have asked in the first place if I knew
anonymous
  • anonymous
\[ \cos^2(2u) = \frac{1+\cos(4u)}{2} \]

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