Question

Integration help!

Answer 1

\[\Huge \int\limits \frac{1}{5 \cos x-12 \sin x}dx\]

Answer 2

divide and multiply by sec^2 x

Answer 3

then go for substitution

Answer 4

Attempt :
\[\Large \int\limits \frac{\sec^2 \frac{x}{2}}{5-5\tan^2 \frac{x}{2}-24 \tan \frac{x}{2}}dx\]
Let tanx/2 = t
Numerator cancels out..

Answer 5

\[\LARGE \frac{-2}{5} \int\limits \frac{dt}{(t+\frac{24}{10})^2-(\frac{26}{10})^2}\]

Answer 6

Now used the formula..
and got
\[\Large \frac{1}{13} \log(\frac{5(5+t)}{1-5t})\]

Answer 7

but the answer is kinda different..
don't know what kind of simplification happened in the answer..
\[\LARGE \frac{1}{13}\log|\sec(x+a)+\tan(x+a)|+C\]
Where
\[\LARGE a= \tan^{-1} \frac{12}{5}\]

Answer 8

have you tried sec x- tan x = 1/ sec x+ tan x and tan 5 + tan t= 5+t/1-5t ?

Answer 9

@hartnn

Answer 10

in denominator,
divide and multiply by 13 to each term

Answer 11

to get the form sin Acos B - cos A sin B

Answer 12

5+t/1-5t remind somewhat about arctan(x+y)/(1-xy) though :o

Answer 13

yeah, but i would start by making the denominator sin (A-B)

Answer 14

it becomes so simple then

Answer 15

how will it get the form of sin AcosB-cosAsinB ?

Answer 16

in denominator,
divide and multiply by 13 to each term

Answer 17

\[\Large 13(\frac{1}{13}-\frac{5}{13}t)\]

Answer 18

i mean in your question!

Answer 19

oh

Answer 20

5/13 cosx - 12/13 sin x

Answer 21

sin(5/13+x)

Answer 22

5/13 = sin t
12/13 = cos t
tan t =...?

Answer 23

how + ?

Answer 24

sorry minus!

Answer 25

so, denom is ?

Answer 26

use "a" instead of t
your answer will match

Answer 27

5/13 = sin a
12/13 = cos a
tan a =...?
a=...

Answer 28

tan a =5/12
a= arctan(5/12)

Answer 29

\[\large \int\limits\limits \frac{1}{5 \cos x-12 \sin x}dx = \int\limits \frac{1}{\sin(\frac{5}{13}-x)}\]
\[\large \int\limits cosec(\frac{5}{13}-x)dx=\log|cosec(\frac{5}{13}-x)-\cot(\frac{5}{13}-x)\] +c

Answer 30

then do this
5/13 = cos a
12/13 = sin a
tan a =...?
a=...
else answer will not match, even though you'll get an equivalent answer

Answer 31

where did the 13 multiplied to the denominator go ?

Answer 32

its outside the integral invisible :D

Answer 33

with tan a =12/5
you will get other form in denominator,
cosAcos B - sin Asin B =...

Answer 34

im confused :( at which step are we? what I did is correct?except 13 outside

Answer 35

what you did is correct, but you will not get a matching answer with what u have

Answer 36

though the answer will be equivalent

Answer 37

umm so where to start from for the required answer?

Answer 38

scratch everything,
do this
5/13 = cos a
12/13 = sin a
tan a =...?
a=...

Answer 39

tan a = 5/12
a= arctan(5/12)

Answer 40

read carefully, i changed it

Answer 41

arctan(5/13) !

Answer 42

what tan a =...?

Answer 43

5/13 = cos a
12/13 = sin a
tan a =12/5

Answer 44

finally! :)
a =..?

Answer 45

arctan(12/5)

Answer 46

so, whats the denominator now ?
(don't forget to include the multiplied 13 too)

Answer 47

latex showoff kar rha hai kya ? :P

Answer 48

\[\Large 13 \int\limits \frac{1}{\cos a \cos x-\sin a \sin x} dx = 13 \int\limits \frac{1}{\cos(a+x})dx\]
\[\Large 13 \int\limits \sec(x+a) dx = 13 \log|\sec(x+a) + \tan (x+a)|\]
where a=arctan(12/5)

Answer 49

:P

Answer 50

abee, 13 was multiplied in denominator...

Answer 51

my bad

Answer 52

chalta hai :|

Answer 53

chal koi nai, any more doubts ?

Answer 54

not on this question :D thanks!

Answer 55

welcome ^_^

Answer 56

did i ever give you this :
http://openstudy.com/users/hartnn#/updates/50960518e4b0d0275a3ccfba

Answer 57

nope,i know most of them but very useful certainly! ill print it out :P its only for indef?

Answer 58

yes, only for indef
mainly i wanted to share tips and shortcuts, but i though to include formulas too..