Question

medium/hard integral question

Answer 1

I'll be over there too.

Answer 2

Is this solvable?

Answer 3

did you try the weistrass substitution
t = tan(x/2) ?

Answer 4

I'm not sure how to approach that : o
Can you help me a little bit?

Answer 5

from my notes :

\(\text{ To integrate} \huge \frac{c\sin\:x+d\cos\:x}{a\sin\:x+b\cos\:x} or \frac{ce^x+d}{ae^x+b}\\ \text{express N as} \large A(D)+B\frac{d}{dx}(D)
\)

Answer 6

N = numerator, D = denominator

Answer 7

forget t= tan (x/2)
i think that will work for
\(\Large \frac{1}{a\sin\:x+b\cos\:x+c} \)
form only

Answer 8

so lets try :
\(169 \sin x = A(5 \sin x+12 \cos x)+B \dfrac{d}{dx} (5 \sin x+12 \cos x)\)

can you find A and B ?

Answer 9

169sin(x) = A+B d/dx (5sinx+12cosx)
A+B (d/dx)= 169sin(x) / (5sinx + 12cosx)

Answer 10

knowing \(5^2 +12^2 =13^2 =169\)
i am tempted to try alternative method.
which i will do LATER.

firstly, dod you know what that d/dx mean ? heard of derivatives ?

Answer 11

yes :)

Answer 12

good, whats
d/dx (5 sin x +12 cos x)
?

Answer 13

please show me a more efficient method! Always look to improve derivative solving skill.

Answer 14

=5cos(x) -12sin(x)

Answer 15

yes, plug that in
\(169 \sin x = A(5 \sin x+12 \cos x)+B \dfrac{d}{dx} (5 \sin x+12 \cos x)\)

and then compare the co-efficients of sin x and cos x

Answer 16

co-efficient of sin x on right side is
5A -12B

and on left it is 129

so, 5A-12B =169
makes sense ?

Answer 17

Yes :)

Answer 18

It is looking a little clearer to me :D

Answer 19

make another equation in A and B
and let me know what values you get for them :)

Answer 20

12 A +5B = 0
5A-12B =169

Answer 21

Sorry I was trying to get the second equation but falied miserably.
Can you show me just the setting up of that equation? Thank you so much >.<

Answer 22

This is really the first time I've seen this type of substitution so. Can I know the name of this method?

Answer 23

Though the part to get A and B after setting up the equation, I understand. :)

Answer 24

\(169 \sin x = A(5 \sin x+12 \cos x)+B \dfrac{d}{dx} (5 \sin x+12 \cos x)\)
\(169 \sin x = A(5 \sin x+12 \cos x)+B (5 \cos x-12 \sin x)\)
\(169 \sin x+0 \cos x = (5A-12B) \sin x + (12A+5B)\cos x \)
\(169 = 5A -12B \\ 0 = 12A+5B\)

Answer 25

I have not given any name to that method yet :P
you can call it <<insert your name here>> method ;)

Answer 26

ohhh : o thank you >.<
You invented this method??? : O That's amazing!!

Answer 27

lol no , i didn't :P
my professor gave us in our notes...idk who invented it:P

Answer 28

The "Got-a-ku" method (Goku) dragon ball :)) <33 sounds like got a clue
idk xD

Answer 29

what can I do next ?

Answer 30

replace numerator with the first equation, then you will see what can be done next :)

Answer 31

\[\frac{ 5A-12B }{ 5\sin(x) + 12\cos(x) }\]

Answer 32

\(\Large \dfrac{ A(5 \sin x+12 \cos x)+B \dfrac{d}{dx} (5 \sin x+12 \cos x)}{5 \sin x+12 \cos x} \\ \Large \dfrac{A(5 \sin x+12 \cos x)}{5 \cos x+12 \cos x}+\dfrac{B(5\sin x-12\sin x)}{5 \sin x+12 \cos x}\)

Answer 33

\(\Large = A+ B\dfrac{(5\sin x-12\sin x)}{5 \sin x+12 \cos x}\)

Answer 34

oh :o
= A+ ( (B(5sinx-12sinx)) / (5sinx + 12cos(x)) )

Answer 35

couldn't you find A,B values?

Answer 36

can I use the 5A-12B=169, or this equation?
Sorry, I'm a rookie at this method so :)

Answer 37

if you haven't noticed the pattern yet :
\[169 \sin x = A(5 \sin x+12 \cos x)+B \dfrac{d}{dx} (5 \sin x+12 \cos x)\]

is exactly same as :

\[\large \text{numerator} = A(\text{denominator })+B \dfrac{d}{dx} (\text{denominator})\]

Our goal is to find the constants \(A\) and \(B\) and replace the \(\large \text{numerator}\) with above pattern.

Answer 38

169=5A−12B

0=12A+5B
its like solving 2 equations in 2 unknowns, just algebra

12A =-5B,
A =(-5/12)B

plug this value of A in 169=5A−12B

Answer 39

169= -25/12 (B) -12B
169= -169/12 (B)
B= -12

Answer 40

correct, what about A ?

Answer 41

A= 5

Answer 42

@ganeshie8 thank you! The goal became much clearer :))

Answer 43

so now you just have to integrate
\(\huge \int 5-12\dfrac{(5\sin x-12\sin x)}{5 \sin x+12 \cos x} dx\)

which is much simpler than original integral

Answer 44

ohhhh I see it now :O

we're going to use ln to solve this?

Answer 45

yes,

(\int \dfrac{f'(x)}{f(x)} dx = ln f(x) +c\)
for the 2nd term

Answer 46

\(\int \dfrac{f'(x)}{f(x)} dx = ln f(x) +c\)

Answer 47

the whole point of doing
\(\large \text{numerator} = A(\text{denominator })+B \dfrac{d}{dx} (\text{denominator})\)
is to get that form
\(\dfrac{f'(x)}{f(x)}\)

Answer 48

I have a question, it's both sin(x) on top.

f'(x) of denominator = 5 cos(x) -12sin(x)

Answer 49

that was just a typo, work this :

\[\huge \int 5-12\dfrac{(5\color{red}{\cos x }-12\sin x)}{5 \sin x+12 \cos x} dx\]

Answer 50

yeah, sorry about that

Answer 51

ok then is 5x -12 ln( 5 sinx +12 cos x) + C

Thank you so so much! @hartnn @ganeshie8
You're amazing >.< <333

Answer 52

I would have no clue of how to work this problem without your Goku method :D

Answer 53

if you still have gas and good with trig, there is another interesting method to work integrals of this form : \(\dfrac{\sin}{ a\sin + b\cos }\)

Answer 54

Goku method sounds nice :3

Answer 55

Please show me :D
Well I'm deflated but this can of juice will pump it up.

Answer 56

xD because I look at hartnn's avi.

Answer 57

maybe il out the solution here, go thru and see if it makes sense

Answer 58

*put

Answer 59

Ok, thank you! x)

Answer 60

the key thing is the denominator again :
\[5\sin x + 12\cos x = \langle 12, 5\rangle \bullet \langle \cos x , \sin x \rangle = \color{red}{13\sin (x-\alpha)} \]
\( \alpha = \arctan(5/12)\)

Answer 61

denominator can be expressed as SINGLE sine function using trig ^^
if you don't like vectors, you may memorize it as formula :)

Answer 62

Once we're comfortable with writing the denominator as a single sine function, below method will make sense :

\[\begin{array}\\

\int \dfrac{169\sin(x)}{5\sin(x) + 12\cos(x)} dx & = \int \dfrac{169\sin(x)}{13\sin(x-\alpha)} dx ~~\color{gray}{ \alpha = \arctan(5/12)} \\
&= 13\int \dfrac{\sin(x-\alpha+\alpha)}{\sin(x-\alpha)} dx \\
&= 13 \left(\int \cos \alpha + \dfrac{\cos(x-\alpha) \sin \alpha}{\sin(x-\alpha)} dx \right) \\
&= 13\cos \alpha \int 1 dx + 13 \sin \alpha \int\cot(x-\alpha) dx \\

\end{array}\]

Answer 63

Sorry it's 2:27 am here so I'll switch to mobile app. :))

Answer 64

You should sleep! that trig identity shows up in differential equations a lot.. go thru when you're free... there is not much in that method except for the unfamiliar trig identity

Answer 65

Wow this method is great! I ask this question too many times, but is there a name for this trig formula? :o

Answer 66

Thank you ganeshie and hartnn. I learnt alot of new and beneficial methods today from you. Hope you have a super great day/night!

Answer 67

have good sleep :) they make this formula the important trig identity in differential equations course :
http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/DiffEq/ConvertingTheForm_asinx.pdf

Answer 68

Just wanted to point out that there's nothing wrong with using the half-angle sub @hartnn mentioned.

Letting \(\tan\dfrac{x}{2}=t\), you have the following reference triangle:

You can work out that
\[\sin x=\frac{2t}{1+t^2}\\
\cos x=\frac{1-t^2}{1+t^2}\\
dt=\frac{1}{2}\sec^2\frac{x}{2}~dx~~\iff~~\frac{2~dt}{1+t^2}=dx\]
So the integral is
\[\int\frac{169\sin x}{5\sin x+12\cos x}~dx=\int\frac{169\left(\dfrac{2t}{1+t^2}\right)}{5\left(\dfrac{2t}{1+t^2}\right)+12\left(\dfrac{1-t^2}{1+t^2}\right)}\frac{2}{1+t^2}~dt\]
which reduces to
\[\int\frac{676t}{(1+t^2)(10t+12-12t^2)}~dt\\
\int\frac{338t}{(1+t^2)(5t+6-6t^2)}~dt\]
Partial fractions from here should be fine.