Question

Calculus II question. Evaluating Integral with trig functions... question posted inside post

Answer 1

\[\int\limits_{\sqrt{\pi/2}}^{\sqrt{\pi}} 7\theta^3 \cos (\theta^2) d \theta \]?

Answer 2

\[ \int\limits\limits_{\sqrt{\pi/2}}^{\sqrt{\pi}} 7\theta^3 \cos (\theta^2) d \theta \]

Answer 3

I think the easiest way is to start with integration by parts

Answer 4

yes. sorry! I am not fast at the equation thing

Answer 5

lol it takes practice :)

Answer 6

substitution could be useful

Answer 7

i would think t=theta^2
dt= 2theta dtheta

Answer 8

or 1/2dt= thetadtheta

Answer 9

idk tho

Answer 10

looks good to me so far

Answer 11

give me a min to write and see what i got plz i am not fast

Answer 12

\[so then I would have \int\limits_{\sqrt{\pi/2}}^{\sqrt{\pi}} 7\theta^3 \cos \theta^2 \theta d \theta\]

Answer 13

is that correct?

Answer 14

or

Answer 15

\[\int\limits_{\sqrt{\pi/2}}^{\sqrt{\pi}} \cos t 1/2 dt\]

Answer 16

meant to throw a 7 theta ^3 before cos :(

Answer 17

\[\int\limits_{\sqrt{\frac{\pi}{2}}}^{\sqrt{\pi}}7 \theta \theta^2 \cos(\theta^2) d \theta\]
so you mean to replace the theta^2 with t
and the theta d theta with 1/2 dt
so you get this:
\[\int\limits_{\frac{\pi}{2}}^{\pi} 7 t \cos(t) \frac{1}{2} dt \]

Answer 18

i also changed the limits

Answer 19

ok.. do i plug into the integral udv = vu- integral vdu now?

Answer 20

or i make the dv v u and du 1st and then plug in right?

Answer 21

you need to decide on which is your u and which is your dv, yes

Answer 22

would dv be 7 cos t dt ?

Answer 23

sounds fine

Answer 24

i guess you will do 1/2 t for the differentiating part

Answer 25

for u?

Answer 26

yah

Answer 27

so then

Answer 28

it doesn't matter where you put the constant though
like you could leave the constant off until the end

Answer 29

dv = 7 cos t dt v= 7 sin t
u = 1/2 t du= 1/2 dt ?

Answer 30

du doesnt look right. what did i do wrong?

Answer 31

im sorry... i really need step by step help with this one! i will fan and reward promise

Answer 32

you didn't substitute t for theta right
myin corrected you but you didn't notice, go back up and look at the substitution she made

Answer 33

i am lost :( already

Answer 34

you had\[t=\theta^2\implies\theta d\theta=\frac12dt\]and your integrand was\[\theta^3\cos(\theta^2)d\theta=\theta^2\cos(\theta^2)\theta d\theta\]try substituting for t again; a littlemore carefully this time

Answer 35

i will start completely over on this question. plz give me a min and hang in there with me

Answer 36

what do i do with the 7 at this point?

Answer 37

nothing?

Answer 38

the 7 can be left out and then factored back in at the end, because\[\int Cf(x)dx=C\int f(x)dx\]if \(C\) is a constant. so you can asically just do \[\int \theta^3\cos(\theta^2)d\theta\]and multiply everything by 7 at the end

Answer 39

ok

Answer 40

basically*
Sometimes you actually want to bring the constant into the integration, because it makes the substitution simpler, but this is not one of those times.

Answer 41

this question is very difficult

Answer 42

just go step-by-step
what was your substitution?

Answer 43

if t= 1/2dt and dt= theta^2 cos theta^2 theta d theta

Answer 44

7 t cos(t) dt

Answer 45

good sorry I was afk

Answer 46

yes, now integrate that by parts\[u=7t\\dv=\cos tdt\]