Question

I tried to integrate this, but, the result is wrong. The correct answer is pi/3. I not find my error, could someone find for me, please? sinx is equals to senx

Answer 1

try to put u=3t and solve again.....

Answer 2

sin(0) = 0

Answer 3

so it would be pi/3 - 0

Answer 4

by the end

Answer 5

that was the only mistake i could find

Answer 6

sin(3pi)=0
cos(3pi)= -1

Answer 7

but has t times in front of cos(3t)

Answer 8

wolfram says your integral is wrong

Answer 9

right,
-t * cos(3t) --> -pi*-1 - 0*1

Answer 10

should be \[\huge \left[\frac 19 \sin (3t) - 3t\cos (3t)\right|_0^\pi\]

Answer 11

I got -pi/3 for the original question.

Answer 12

The integration looks ok. The mistake is in evaluating the limits.

Answer 13

http://www.wolframalpha.com/input/?i=int+%28tsin+%283t%29%29dt

Answer 14

unless i made a syntax error...the integration is wrong

Answer 15

Oops no I didn't I got pi/3. Shall write working now.

Answer 16

@lgbasallote The integration is ok.

Answer 17

so wolfram is wrong?

Answer 18

wolfram is correct.

Answer 19

hmm yeah seems they're the same

Answer 20

@jpsmarinho redo substituting in your limits
for the first term, use cos(0)=1 cos(3pi)= -1
for the 2nd term use sin(0)=0, sin(3pi)=0

Answer 21

I=int(uv')=uv-int(u'v). Have u=t, =>u'=1. v'=sin(3t) => v=-1/3 cos(3t)
I=uv-int(u'v)=-t/3 cos(3t) -int(-1/3 cos(3t))
I=-t/3 cos(3t) +1/9sin(3t)
I=pi/3

Answer 22

Wolfram says this is correct. http://www.wolframalpha.com/input/?i=int+%28tsin+%283t%29%29dt+between+0+and+pi

Answer 23

pardon, i didn't understand what's the wrong! I redo some part of the exercise and the answer is the same!

Answer 24

your first term is correct, but your 2nd should be zero.
\[\frac{ -tcos(3t) }{ 3 } \bigg|_{0}^{\pi}+\frac{\sin(3t)}{9}\bigg|_{0}^{\pi}\]
\[\frac{ -\pi \cos(3\pi) }{ 3 }-\frac{ -0 \cdot \cos(0) }{ 3 }+\frac{\sin(3\pi)}{9}-\frac{\sin(0)}{9}\]
the last 3 terms are all zero

Answer 25

@phi thanks!! I don't know why my mind blocked to it. ^^