Question

Use the indicated substitution to evaluate the integral :

The Integral of (1-cos(t/2))^2 x (sin(t/2)) dt u= 1-cos(t/2)

Answer 1

du = sin(t/2)/2 dt

so integral becomes (u^2) 2du = 2u^3 /3 + C

just sub u= 1-cos(t/2) back

Answer 2

Where does the division ( the sin(t/2) / 2 come from.

Answer 3

differentiating

u= 1-cos(t/2)

Answer 4

I think you are thinking of something else. You would then have to solve for dt first.... i.e. dt= 2du/sin(t/2) and then I would plug it back in. But thanks.

Answer 5

I'm correct. I've done this plenty of times.

Answer 6

I'll fully type it out with latex if you don't know what I'm saying.

Answer 7

Hahaha you got the wrong answer though. Its chill. BC Calc sucks .

Answer 8

\[\int\limits (1 - \cos ({t \over 2}))^2 \times \sin({t \over 2}) dt\]

now, firstly:

\[u = 1 - \cos({t \over 2}) \rightarrow du = {1 \over 2}\sin({t \over 2}) dt \rightarrow \sin({t \over 2}) dt = 2du\]

now, replace sin(t/2)dt with 2 du and 1- cos(t/2) with u

the integral becomes:

\[\int\limits 2u^2 du = {2 \over 3} u^3 + C = {2 \over 3}(1-\cos({t \over 2}))^3 + C\]

Answer 9

This answer is correct.

Answer 10

k that made a lot more sense.

Answer 11

Actually you need to pull it to .....^3/2 i.e. anti derivative formula of: \[(X^(n+1))/ (n+1) \]

Answer 12

I did that.

Answer 13

You left it as cubed rather than 3/2

Answer 14

why on earth would it be 3/2?

Answer 15

\[\int\limits u^2 = {u^3 \over 3} +C\]

Answer 16

Mentor teaching the mentee huh.

When you take the Anti-Derivative of
\[\sqrt{\tan x}\]
The formula that I just showed you can be applied.
So \[\sqrt{\tan x}\]
is technically \[\tan^.5\] Correct?

Putting it into the equation you would get that result.

Answer 17

Of 2/3 tan ^ 3/2

Answer 18

You're teaching me how to laugh at you, you're not teaching me any math. THERES NO SQUARE ROOT IN YOUR QUESTION.

Answer 19

whoopsy looking at the wrong equations. Don't get butt hurt guy.

Answer 20

don't try and act smart.

Answer 21

Soooooooo Butt hurt. My GPA laughs at you.

Answer 22

Just try and get through high school.

Answer 23

oh, and by the way, if you can't do an integral as simple as this, I highly doubt your GPA is even able to laugh.

Answer 24

4.0 Unweighted. And its sad to see you just learning this in College but I don't judge.

Answer 25

I didn't learn this in college. I learned this at school. The fact that I can do this while I'm in university says nothing about when I learnt this. I'm not a math major either.