Question

integrate 7x cos^4(x^2) dx

Answer 1

Try \[
u=x^2\quad du=2x\;dx
\]

Answer 2

Also \[
\cos^2(u) = \frac{1+\cos(2u)}{2}
\]

Answer 3

The only question is, how badly do you want it?

Answer 4

The 7x is confusing me

Answer 5

My instinct says to solve it like an integration by parts

Answer 6

Okay, here is some basic algebra: \(7x=3.5\cdot 2x\)

Answer 7

\[
7x = \frac 7 2 \cdot 2x
\]

Answer 8

\[
\int 7x \cos^4(x^2) dx = \int \frac 7 2 \cos^4(x^2)\;2xdx
\]

Answer 9

It's not like there is any obvious pattern here or anything.\[
u = x^2\quad du=2xdx
\]

Answer 10

ah, okay

Answer 11

\[\int\limits_{}^{}\frac{ 7 }{ 2 }\cos ^{4}(u) du\]

Answer 12

?

Answer 13

\(\color{blue}{\text{Originally Posted by}}\) @wio
Also \[
\cos^2(u) = \frac{1+\cos(2u)}{2}
\]
\(\color{blue}{\text{End of Quote}}\)

Answer 14

so you plug that in for cos and square the whole formula?

Answer 15

Yes.

Answer 16

and bring out the 7/2 constant

Answer 17

Sure.

Answer 18

\[\frac{ 7 }{ 2 }\int\limits_{}^{}(\frac{ 1+\cos(2u) }{ 1 })^{2}u du\]

Answer 19

Not \(udu\), just \(du\)

Answer 20

2 in the denominator

Answer 21

Now use foil.

Answer 22

what happened to the x^2 in cos^4x^2?

Answer 23

\(u=x^2\)

Answer 24

I'm very confused for some reason. Is the half angle formula squared because of the x^2 or the cos^4?

Answer 25

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

Answer 26

It's because the \(\cos\) is squared.

Answer 27

oh, I think it just clicked

Answer 28

okay

Answer 29

\[
\cos^4(u) = [\cos^2(u)]^2 = \left[\frac{1+\cos(2u)}{2}\right]^2
\]

Answer 30

so now we're multiplying it out

Answer 31

\[\frac{ 7 }{ 2 }\int\limits_{}^{}\frac{ 1+2\cos(2u)+\cos ^{2(2u)} }{ 4} du\]

Answer 32

not an exponent on the "2u" obviously

Answer 33

Pull out the 4 and use the half angle again on the squared cosine.
Then it will be simple.

Answer 34

okay, I got this

Answer 35

I doubt I did this right\[\frac{ 14 }{ 16 }\int\limits_{}^{}2+2\cos(2u) du\]

Answer 36

Show your work.

Answer 37

\[\frac{ 7 }{ 8 }\int\limits_{}^{} 1+2\cos(2u)+(\frac{ 1+\cos(2u) }{ 2 }) du\]

Answer 38

should be \(4u\)

Answer 39

on the first cos?

Answer 40

Okay, which one do you think I was talking about?

Answer 41

the first one. I didn't have to do anything with the second besides plug in the formula

Answer 42

nevermind

Answer 43

I dont see why either one would be 4u

Answer 44

and I wouldn't have asked in the first place if I knew

Answer 45

\[
\cos^2(2u) = \frac{1+\cos(4u)}{2}
\]