Question

integral of sin^4 (6x) dx

Answer 1

Take the integral:
integral sin^4(6 x) dx
For the integrand sin^4(6 x), substitute u = 6 x and du = 6 dx:
= 1/6 integral sin^4(u) du
Use the reduction formula, integral sin^m(u) du = -(cos(u) sin^(m-1)(u))/m + (m-1)/m integral sin^(m-2)(u) du, where m = 4:
= 1/8 integral sin^2(u) du-1/24 sin^3(u) cos(u)
Write sin^2(u) as 1/2-1/2 cos(2 u):
= 1/8 integral (1/2-1/2 cos(2 u)) du-1/24 sin^3(u) cos(u)
Integrate the sum term by term and factor out constants:
= -1/24 sin^3(u) cos(u)+1/8 integral 1/2 du-1/16 integral cos(2 u) du
For the integrand cos(2 u), substitute s = 2 u and ds = 2 du:
= -1/32 integral cos(s) ds-1/24 sin^3(u) cos(u)+1/8 integral 1/2 du
The integral of cos(s) is sin(s):
= -(sin(s))/32+1/8 integral 1/2 du-1/24 sin^3(u) cos(u)
The integral of 1/2 is u/2:
= -(sin(s))/32+u/16-1/24 sin^3(u) cos(u)+constant
Substitute back for s = 2 u:
= u/16-1/32 sin(2 u)-1/24 sin^3(u) cos(u)+constant
Substitute back for u = 6 x:
= (3 x)/8-1/32 sin(12 x)-1/24 sin^3(6 x) cos(6 x)+constant
Which is equal to:
Answer: |
| = 1/192 (72 x-8 sin(12 x)+sin(24 x))+constant

Answer 2

As an alternative, you can rewrite the intregrand with the half-angle identity:
\[\sin^2\theta=\frac{1}{2}\left(1-\cos(2\theta)\right)\]
So, you can rewrite the integrand:
\[\begin{align*}\sin^4(6x)&=\left[\frac{1}{2}\left(1-\cos^2(12x)\right)\right]^2\\
&=\frac{1}{4}\left(1-\cos(12x)\right)^2\\
&=\frac{1}{4}\left(1-2\cos(12x)+\cos^2(12x)\right)
\end{align*}\]
And using the identity involving cosine:
\[\cos^2\theta=\frac{1}{2}(1+\cos(2\theta))\]
So,
\[\begin{align*}\sin^4(6x)&=\frac{1}{4}\left(1-2\cos(12x)+\frac{1}{2}\left(1+\cos(24x)\right)\right)\\
&=\frac{1}{4}\left(\frac{3}{2}-2\cos(12x)+\frac{1}{2}\cos(24x)\right)\\
&=\frac{3}{8}-\frac{1}{2}\cos(12x)+\frac{1}{8}\cos(24x)
\end{align*}\]
Then term-by-term integration can be easily done.

Answer 3

I'm sorry i should have told you that my professor does not want us to use the formula, this is what i saw on wolf but I'm trying to integrate it out all without using the formula... so far I'm down to 1/24[u-(2sin(2u)/2)] + integral cos^2 (2u) du

Answer 4

Does he just want you to use your calculator?

Answer 5

no no no, this is all done analytically - @SithsAndGiggles this is partially of what i have started to do, I started out using u-sub for 6x then used half-angle identity.

Answer 6

@sjerman1, did you check my solution, I don't think I use half-angle.

Answer 7

\[\int\limits \sin^4 (6x) dx \]
u=6x du=6dx -> 1/6du=dx
\[1/6 \int\limits \sin^4 (u) du\]
\[\sin^2 u = \frac{ 1-\cos(2u) }{ 2 } \therefore \sin^4 u = [\frac{ (1-\cos(2u) }{ 2 }]^2 = \frac{ 1-2\cos(2u)+\cos^2 (2u) }{ 4 }\]
\[\int\limits\limits \sin^4 (6x) dx = \frac{ 1 }{ 6 } \int\limits \frac{ 1 }{ 4 } (1-2\cos(2u)+\cos^2(2u)) du\]
\[\frac{ 1 }{ 6 }\times \frac{ 1 }{ 4 } = \frac{ 1 }{ 24 }[u-\frac{ 2\sin(2u) }{ 2 }] + \int\limits \cos^2 (2u) du\]

Answer 8

@Zelda I noticed that you didnt use the half angle formula, but my professor does not want me using the formulas such as the reduction formula that you pulled from wolframalpha

Answer 9

Fair enough. But gotta love wolfram alpha. I hate integrating myself >.>

Answer 10

Which formulas are you allowed and not allowed to use?

Answer 11

trig formula, half-angle formula and trig identities. nothing that is like a plug and play formula such as the reduction formula

Answer 12

\[16×14=124[u−2\sin(2u)2]+∫\cos2(2u)du\]
\[\int\limits \cos^2 (2u) du = \int\limits \frac{ 1 }{ 2 } + \frac{ 1 }{ 2 }\cos4u du\]
\[\frac{ 1 }{ 24 }[\frac{ 3 }{ 2 }u - \frac{ 2\sin(2u) }{ 2 } +\frac{ 1 }{ 2 } \frac{ \sin(4u) }{ 4 }+c]\]
replace 6x for u
\[\frac{ 1 }{ 24 }[9x-\frac{ 2\sin(12x) }{ 2 }+\frac{ 1 }{ 8 }\sin(24x)+c]\]
there we go, now we are done...