Question

help please
evaluate the integral

Answer 1

't' and 'x'? And is the intergral 'dx or dt?

Answer 2

oh wait dt sorry

Answer 3

https://youtu.be/xhy7dXWjpAA

Answer 4

Substitute \[u=5x\]

Answer 5

\[\int\limits_{}^{} t^2 \sin (5t)\]

Answer 6

Your last question was like this. You'll have to do int. by parts a couple times.

Answer 7

yes , im still confused bc i know it has to follow this form LIATE ?

Answer 8

It's almost always easiest to pick the function which eventually will differentiate to become 0... which of sin5t or t^2 will eventually have a zero derivative?

Answer 9

Firstly you're still mixing up x and t. Secondly, you didn't pick a dv like you're supposed to.

Answer 10

u substitution wont work here

Answer 11

aaahhh true lol

Answer 12

im so confused though hmmm let me fix this real quick

Answer 13

u= t^2 dv = sin (5t) dt
du= 2t dt v= - 1/5 cos (5t)

Answer 14

is this right ?

Answer 15

Yes

Answer 16

now put it together

Answer 17

okay so then


\[-t^2 * \frac{ 1 }{ 5 } \cos (5t) - (- \frac{ 1 }{ 5 }) \int\limits_{}^{}\cos (5t) * 2t \]

Answer 18

so then we integrate again ?

Answer 19

\[-\frac{ 1 }{ 5 }Cos(5t)~t^{2}+\int\limits (\frac{ 1 }{ 5 }Cos(5t)~2t)~dt\]

Answer 20

Now use int. by parts again, on that remaining integral.

Answer 21

yep you need to integrate again,

Answer 22

evaluate the second integral the same way you did the first one @marcelie

Answer 23

aah i think i made a mistake somewhere \[-\frac{ 1 }{ 5 }\cos (5t) t^2 +\frac{ 1 }{ 5 } ( 2t*\frac{ 1 }{ 5 } *- \sin (5t) *2 )\]

Answer 24

I'd recommend you show the work for the separate integral separately.

Answer 25

u1= 2t dv : cos (5t)

du 1 = 2 dt v= - 1/5 sin (5t)

Answer 26

Int. of cosine is positive sine.

Answer 27

oh riight lol okay so then

Answer 28

you there D:

Answer 29

Well i was waiting for you to continue...