Question

find the definite integral ∫(sin^3)6x(cos^4)6xdx

Answer 1

No domain, indefinite...

Answer 2

Go to www.WolframAlpha.com
Paste in the following to the right of the equal sign:
Integrate[Sin[x]^3*6*x*Cos[x]^4*(6*x), x]
When the answer is presented, click on "show steps"
The integration procedure is very lengthy for this problem.
A solution copy from my own Mathematica 8 Home Edition is attached.

Answer 3

yeah, I'm a little lost by looking at it.

Answer 4

Maybe u could get away with rewriting the cos^4 as (1-sins^2)^2?

Answer 5

\[\int\limits_{}^{}\sin^2(6x)*\cos^4(6x)*\sin(6x)dx\]
\[\int\limits_{}^{}(1-\cos^2(6x))\cos^4(6x)*\sin(6x)dx\]
\[\int\limits_{}^{}(\cos^4(6x)*\sin(6x)-\cos^2(6x)\cos^4(6x)*\sin(6x))dx\]
\[\int\limits_{}^{}\cos^4(6x)*\sin(6x) dx-\int\limits_{}^{}\cos^6(6x)*\sin(6x)dx\]

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Let \[u=\cos(6x)=> \frac{du}{dx}=-6\sin(6x) => du=-6\sin(6x) dx\]
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\[\frac{-1}{6}\int\limits_{}^{}u^4du-\frac{-1}{6}\int\limits_{}^{}u^6du\]

Answer 6

\[\frac{-1}{6}*\frac{u^5}{5}+\frac{1}{6}*\frac{u^7}{7}+C\]
\[\frac{-1}{6}*\frac{\cos^5(6x)}{5}+\frac{1}{6}*\frac{\cos^7(6x)}{7}+C\]

Answer 7

\[-\frac{\cos^5(6x)}{30}+\frac{\cos^7(6x)}{42}+C\]

Answer 8

Where's the sin 6x come from?

Answer 9

sin^3(6x)=sin^2(6x)sin(6x)

Answer 10

Ah!

Answer 11

smart

Answer 12

lol, I'm going to start paying you, myininaya

Answer 13

CHECKING:
\[[\frac{-\cos^5(6x)}{30}+\frac{\cos^7(6x)}{42}+C]'\]
\[\frac{-1}{30}*5*\cos^4(6x)*6*(-\sin(6x))+\frac{1}{42}*7*\cos^6(6x)*6*(-\sin(6x))+0\]
\[\cos^4(6x)\sin(6x)-\cos^6(6x)\sin(6x)=\sin(6x)(\cos^4(6x)-\cos^6(6x))\]
\[=\cos^4(6x)\sin(6x)(1-\cos^2(6x))=\cos^4(6x)\sin(6x)\sin^2(6x)\]
\[=\cos^4(6x)\sin^3(6x)\]

Answer 14

Isn't it easier to do what I suggested and just integrate

Sin^7(6x) -2Sin^5(6x) +Sin^7(6x) ?

Answer 15

lol

Answer 16

Sorry, should be Sin^7(6x) -2Sin^5(6x) +Sin^3(6x)

Answer 17

i always try to play with the odd power if there is one

Answer 18

how did you get that estudier?

Answer 19

What I said in my first post use Cos^2 + Sin^2 = 1 to replace Cos^4

Answer 20

And multiply it out, of course....

Answer 21

could either of you extraordinary gentlemen perhaps help me solve \[\int\limits_{}^{} \sec^57xtan7xdx\]

Answer 22

it might be okay but I think you will still need to us identity to integrate you know so we get du part somewhere

Answer 23

That's a new question....put it up.

Answer 24

There's a Tans Secs identity...

Answer 25

1 + Tan^2 = Sec^2

Answer 26

Remember:
\[\sin^2x+\cos^2x=1\] this next item can be obtain by dividing the first item by cos^2x
\[\tan^2x+1=\sec^2x\]
\[\int\limits_{}^{}\sec^2(7x)\tan(7x)\sec^2(7x)dx\]
and also remember that (tanx)'=sec^2x
\[\int\limits_{}^{}(\tan^2(7x)+1)\tan(7x)\sec^2(7x)dx\]
\[\int\limits_{}^{}\tan^2(7x)\tan(7x)\sec^2(7x)dx+\int\limits_{}^{}\tan(7x)\sec^2(7x)dx\]
\[\int\limits_{}^{}\tan^3(6x)\sec^2dx+\int\limits_{}^{}\tan(7x)\sec^2(7x)dx\]

Answer 27

oops we had sec^5(7x) not sec^4(7x)

Answer 28

Details...

Answer 29

\[\int\limits_{}^{}\sec^4(7x)\sec(7x)\tan(7x)dx\]
now what is the derivative of secx?
this one was way easier than the first

Answer 30

\[u=\sec(7x)=> du=7\sec(7x)\tan(7x)dx\]

Answer 31

\[\frac{1}{7}\int\limits_{}^{}u^4 du\]

Answer 32

\[\frac{1}{7}*\frac{u^5}{5}+C=\frac{1}{35}u^5+C=\frac{1}{35}\sec^5(7x)+C\]

Answer 33

why not make u=7x? same thing, but wouldn't it be faster?

Answer 34

but again, you have my eternal gratitude

Answer 35

if you let u=7x, you would have to make another substitution

Answer 36

oh, right.