Question

NOT REALLY A BASIC INTEGRAL CALCULUS PROBLEM

Answer 1

Are you working out all your homework problems? haha

Answer 2

lol

Answer 3

\[\int\limits sinx (2-3cosx)^2dx\]

Answer 4

@Jhannybean actually im trying to practice most of the problems of my book. if i cant answer it on my own. ill seek help xD

Answer 5

this is the last one. and then next ill be studying physics

Answer 6

Nice, nice.

Answer 7

So this is most likely another u-sub

Answer 8

Can you guess what you'd be substituting?

Answer 9

U = sinx DU = cosx?

Answer 10

No that would be too complicated.

Answer 11

Try \[u= \cos(x) ~, ~ du=-sin(x)dx\]

Answer 12

woops. ok

Answer 13

So what would you get?

Answer 14

\[- \int\limits (4 -12u-9u^2)du\]

Answer 15

im gonna distribute them right?

Answer 16

\[-4cosx + 6\cos^2x - 3\cos^3x + c\]

Answer 17

Did you simplify that right? i'm getting something different.

Answer 18

kinda, whats your answer?

Answer 19

i distributed the negative sign outside

Answer 20

Oh I see what you did.

Answer 21

xD

Answer 22

should i go with my answer?

Answer 23

You got up to here: \[- \int\limits (4 -12u-9u^2)du\] so just add in u's :)

Answer 24

Yeah I messed up in my post.

Answer 25

i just found another way to do it as well, double substituoin, haha, just trying out different methods here, sorry

Answer 26

xD

Answer 27

thanks as always

Answer 28

But continuing from the method we're working on, we have \[\rightarrow\int sinx (2-3cosx)^2dx\]\[\text{let} ~ u = \cos(x) ~, ~ du = -\sin(x)dx\]\[=\int(2-3u)^2du\]\[=\int (4-12u-9u^2)du\]\[=4u-6u^2-3u^3\] And you just plug in your u value.

Answer 29

Ok, the other COOLER method, hahaha.

Answer 30

Let's take it from here:\[\rightarrow \text{let} ~ u = \cos(x) ~, ~ du = -\sin(x)dx\]\[=\int(2-3u)^2du\]\[\text{let w}=2-3u~ , ~ dw = 3dw\]\[=\frac{1}{3}\int w^2dw\]\[=\frac{1}{3}\cdot \frac{w^3}{3}\]\[=\frac{(2-3u)^3}{9}\]\[=\frac{4-12u+9u^2}{9}\]\[=\frac{1}{9} (4-12(\cos(x))+9(\cos^2(x)))\]

Answer 31

I see

Answer 32

Omg no.

Answer 33

I squared it and it's supposed to be cubed.

Answer 34

Let's just leave it at \[\frac{1}{9}(2-3(\cos(x)))^3\] Hahaha

Answer 35

xDD

Answer 36

from this step \[\rightarrow \frac{(2-3u)^3}{9}\]

Answer 37

Soorryy. tiiired! x_x

Answer 38

its fine. xD

Answer 39

But yeah, i tried showing you two methods of evaluating it xD

Well...more like 1 method, 2 steps. (1 u-sub or a double u-sub) xD

Answer 40

well i learned a new method xD

Answer 41

Woot.

Answer 42

hey thanks again

Answer 43

No problem :)