Question

what is the integral of [ t^9 sin(1-t^5) dt ]

Answer 1

this is super easy..for sat

Answer 2

lol hell no

Answer 3

I'd use by parts

Answer 4

i was thinking u - sub then parts, but perhaps best is to use the "subtraction" formula for \(\sin(1-t^5)\)

Answer 5

Yeah you would need a u sub at first though

Answer 6

that would be messy also

Answer 7

think it might be easier to write it as \[t^9\sin(1)\cos(t^5)-t^9\cos(1)\sin(t^5)\]

Answer 8

if you start with \(u=1-t^5\) you have to solve for \(t^9\) in terms of \(u\) which is going to be a mess

Answer 9

I think it's messy either way

Answer 10

Actually no I think yourmethod is better

Answer 11

no this way you have \(u=t^5\) so first one is \[\frac{\sin(1)}{5}\int u\cos(u)du\]

Answer 12

i should shut up i hate this stuff

Answer 13

u=t^5 is a good sub

Answer 14

but it will still be messy xD

Answer 15

\[\int\limits \frac{ 1 }{ 5 } usin(1-u)du\]

Answer 16

Haha I think you would do another u sub then by parts omg

Answer 17

yeah that is why it is probably easiest to use subtraction angle formula first

Answer 18

Yeah haha, ok op you have the techniques, good luck!

Answer 19

\[\sin(1)\int t^9\cos(t^5)dt-\cos(1)\int t^9\sin(t^5)dt\]is where i would start

Answer 20

U-sub into parts doesn't seem that bad :p
Only requires a single parts application because of how t^9 breaks down.

\(\large\rm u=1-t^5\qquad\qquad\to\qquad\qquad 1-u=t^5\)
\(\large\rm -\frac{1}{5}du=t^4dt\)

And use those three pieces to give you\[\large\rm -\frac15\int\limits(1-u)\sin u~du\]
And then parts, bam bam bam :D

Answer 21

@dramirez9 what do you think? :)
Figure this one out yet?

Answer 22

@zepdrix THANKS!!!!!! it was becoming a nightmare using substitution by parts.

Answer 23

Got it? cool cool

Answer 24

\c:/