Question

To sirm3d
#28 plz :)

Answer 1

@sirm3d

Answer 2

implicit differentiation.

Answer 3

for implicit, do u need to do it for both x and y

Answer 4

\[\large x^3 \cos y +e^x=1+e\]\[\large 3x^2 \cos y + x^3 (-\sin y )\frac{ dy }{ dx }+e^x =0 + 0\]

Answer 5

whenever you differentiate an expression with y, don't forget the rule to insert dy/dx

Answer 6

ahh ok, let me try that out :D just need to isolate dy/dx
and then sub in both x and y to solve for dy/dx

Answer 7

why don't you sub first, isolate later.

Answer 8

is it easier :D ok, i will try that haha
btw, did u learn statistics?

Answer 9

oh yea it is easier, cuz cos (pi/2) = 0 , and ans is e :D

Answer 10

it's not my forte, but i can answer stat questions too.

Answer 11

\[\Huge \color{red} \checkmark \]

Answer 12

either taking it in summer or next year :D

Answer 13

I should get #29, wait LOL.. i've been doing so many trig sub..how can i not get this lol,, wait >.>

Answer 14

i'll be around. it not like the world is going to end this Dec 21, 2012. LOL

Answer 15

lol i hope it ends one day earler, never need to see my final mark :D

Answer 16

and u said if trig is even, u use by parts right

Answer 17

odd is subs

Answer 18

that works only for sine and cosine. tangent and secant behave differently.

Answer 19

how do they behave? cuz now i'm stuck with a tan ^4 (X)

Answer 20

x=3sec theta ?

Answer 21

yea

Answer 22

\[\large \int\limits \frac{3 \tan \theta}{(3 \sec \theta)^3}3\sec \theta \tan \theta d \theta=(1/3)\int\limits \frac{ \tan^2 \theta }{ \sec^2 \theta } d \theta\]

Answer 23

it looks like cos^2 theta after simplification

Answer 24

tan^2 theta x 1/ sec ^2 theta = tan^4 theta isnt it

Answer 25

tan = sin / cos
so tan^2/sec^2 = (sin^2)/(cos^2 sec^2) = sin^2, not cos ^2

Answer 26

(cos)(sec) = 1

Answer 27

\[\frac{ \sin^2 x }{ \cos^2 x } * \frac{ \sin^2 x }{ \cos^2 x }\]

Answer 28

i thought it's like that..

Answer 29

1/sec = cos

Answer 30

o..

Answer 31

so sec = cos / sin
but 1 / sec is not sin/ cos

Answer 32

sec = 1/cos, tan = sin/cos = sec/csc
csc = 1/sin, cot = cos/sin = csc/sec

Answer 33

tan/sec = tan * (1 / sec) = (sin/cos)(cos)=sin

Answer 34

omg. how can i get that wrong..
that trig sub confused me.. cuz i go directly from sec to tan..and now messed the easy stuff

Answer 35

sec^2 = tan^2 + 1

Answer 36

my opening sentance was i must know how to do this because i have done so many trig subs
my ending sentance is i have done too many trig subs, thinking tan <-> sec not tan <-> cot
@_@

lol but i remember it now :D

Answer 37

the burden of remembering all the prerequisites makes calculus more difficult.

Answer 38

so now since sin is even, then use sub?

Answer 39

nvm..

Answer 40

no derivative present..

Answer 41

even ---> half-angle
odd ----> sub

Answer 42

then what's by parts? even or odd?

Answer 43

you noticed no derivative. that means you're learning.

Answer 44

even ---> half-angle

Answer 45

cos^2 x = (1 + cos 2x)/2
sin^2 x = (1 - cos 2x)/2

Answer 46

yea, learned alot from u :D

Answer 47

\[\frac{ 1 }{ 6 }\sec ^{-1}(\frac{ x }{ 3 }) -\frac{ \sqrt{x^2-9} }{ x } \]

how come i got this..

Answer 48

\[\int\limits=\frac{ \theta }{ 2 }-\frac{ \sin 2 \theta }{ 4 }+C\]

Answer 49

sin 2theta = 2 sin theta cos theta

Answer 50

oh..u cant directly sub it in

Answer 51

oh, there's stilla 1/3 multiplier that i missed. just insert it later.

Answer 52

x=3sec theta, not x = 3sec 2theta.

Answer 53

okay :D got the ans . i think is also save to conclude the ans is D by seeing a 1/6

Answer 54

right.

Answer 55

i will try #30 first, i really think i will get it !! it's partial fraction haha

Answer 56

okay. then i'll teach you another trick in partial fractions.

Answer 57

yes, plz :D

Answer 58

transfer all your knowledge to me haha =P

Answer 59

i got A= 55/9... do u think ths number looks a bit weird..

Answer 60

it is. its not even among the choices.

Answer 61

but the A value does not necessarily reflect in the final ans isnt it?

Answer 62

but yea.. A= 3

Answer 63

and it does appear in the ans @@

Answer 64

that means we can eliminate the option choice to D or E ?

Answer 65

now you're beating the choices.

Answer 66

C also has a 3, so you can't count it out.

Answer 67

but i only have A B and C , that means i should have only 3 parts, and C and E option have 4, save to conclude it's D?

Answer 68

yes, by number of terms, it's D. BUT...
the coefficient is (3/2), not 3.

Answer 69

i think i just did it wrong..i got B as 0..

Answer 70

oh yea, i know it cant be 0 because ther eis a (x-2)^2 < which will become x-2 after integration

Answer 71

@_@

Answer 72

wait i know why..sorry lol..keep making mistakes ..

Answer 73

yea ts C!

Answer 74

A= 3 B=-1 C= 2

Answer 75

it's 3/2 because ln2 +1 < the 2 is taken out becoming 3/2

Answer 76

Good. you spotted the mistake in choice D.

Answer 77

my mistake is really stupid..i copied the question wrong lol

Answer 78

but it is still the best answer, isn't it.

Answer 79

yes :D finally done my exam practice ques and practice exam, but still feel so unprepared :/

Answer 80

care to try this partial fraction problem? \[\large \int\limits \frac{7x^2-32x+31}{(x-1)(x-2)(x-3)}dx\]

Answer 81

ok :D

Answer 82

or this more challenging, \[\large \int\limits \frac{2x^3+3x-1}{(x-1)^4}dx\]

Answer 83

i will do all haha

Answer 84

3 ln (x-1) + 5 ln (x-2) - ln (x-3)

Answer 85

solution to 1st problem is right.

Answer 86

you should do (x^2 -1)^4 in the denominator lol

Answer 87

it's only (x-1)^4. hehehe.

Answer 88

let's write \[\frac{2x^3 + 3x - 1}{(x-1)^4}=\frac{ A }{ x-1 }+\frac{ B }{ (x-1)^2 }+\frac{ C }{ (x-1)^3 }+\frac{ D }{ (x-1)^4 }\]

Answer 89

wait!

Answer 90

I did it , wait

Answer 91

i was saying u can try x^2 -1 while u are waiting for me :D

Answer 92

I got 2 ln (x-1) + 3/ (x-2)^2 + 1/ (x-1)^3

Answer 93

1/(x^2 - 1) ?, there's a standard integration formula for that

Answer 94

i got b as 0 again..whydo i always get B=0..

Answer 95

really, cuz x^2 u need to do Ax+B thing.. which will turn into a mess

Answer 96

\[\large \int\limits \frac{ Ax+B }{ x^2-a^2 }dx=\frac{ A }{ 2 }\ln \left| x^2-a^2 \right|+\frac{ B }{ 2a }\ln \left| \frac{ x-a }{ x+a } \right|+C\]

Answer 97

ohh.. i dont think i can memorize this formula for now.. but later for sure :P