Question

Integral of (√tanx+√cotx)dx where x€(π,3π/2)

Answer 1

You may split them into two and try, \[\large \int\limits_{\pi}^{3 \pi/2} \sqrt{tanx}dx + \int\limits_{\pi}^{3 \pi/2} \sqrt{cotx}dx\]

Answer 2

It's d question of indefinite integration ....the value of x is just given to ensure that there may b no negative sign under the radical...

Answer 3

Actually this question is in parts in first part they ask tge same ques when x ranges from (0,π/2) i did that...bt this part is what i m not getting

Answer 4

In any case, you may apply the same method, though it will be messy. It's very late here maybe @ganeshie8 @ikram002p may help you with better techniques, good luck.

Answer 5

all i can see is doing this sub
tan x=u^2 maybe

Answer 6

I'll give you HINT

Answer 7

\[\large \bf \int\limits \sqrt{tanx}\]
\[\large \bf Put~tanx=t^2\]
\[\large \bf x=\tan^{-1}t^2\]
\[\large \bf dx=\frac{1}{1+t^4}2t dt\]

Answer 8

then

Answer 9

Yaa....sure @mayankdevnani
@ikram002p i did it by opening up tanx and cotx in terms of sinx n cos x...

Answer 10

\[\sqrt{\tan x} + \sqrt{\cot x} = \dfrac{\sin x+\cos x}{\sqrt{\sin x \cos x}} = \dfrac{\sqrt{2}(\sin x-\cos x)'}{\sqrt{1-(\sin x -\cos x)^2}}=\cdots \]

Answer 11

\[\large \bf \int\limits t.\frac{1}{1+t^4}.2t.dt=\int\limits \frac{2t^2}{1+t^4} dt\]

Answer 12

Now we can substitute sinx-cosx as t....then cosx +sin x dx will be dt...

Answer 13

By opening tax into sinx and cosx , it become so MESSY

Answer 14

N d denominator term will be √1-t^2...

Answer 15

N we know d integration of dt/√1-t^2 which is arc sint

Answer 16

Yes

\[\int\dfrac{dt}{\sqrt{1-t^2}} = \arcsin(t) + C\]

Answer 17

Now here comes the most important step

Answer 18

Bt the ans is giving a -ve sign also in front of arcsint...That's what i m nlt getting...

Answer 19

substitute the value of t and take the bounds

Answer 20

Not*

Answer 21

lal i feel wetawded

Answer 22

\[\large \bf \int\limits \frac{(1+t^2)+(t^2-1)}{1+t^4} dt\]
\[\large \bf \int\limits \frac{1+t^2}{1+t^4}dt+\int\limits \frac{t^2-1}{1+t^4}dt\]

Answer 23

Now you can solve it easily !

Answer 24

bt then also i feel that ans will be arcsin(sinx-cosx)...Isn't it @ganeshie8 ?

Answer 25

one sec..

Answer 26

\[\large \bf \int\limits \frac{1+x^2}{1+x^4}dx=\int\limits \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2}\]
\[\large \bf put~ x-\frac{1}{x}=t\]
\[\large \bf \int\limits \frac{dt}{2+t^2}\]

Answer 27

^^ Here is the way to solve it^^^

Answer 28

Here is my method(one of the short method)

Answer 29

that is nice :)
my method is not giving correct answer, see anything wrong ?

Answer 30

Ryt.....it is not giving us the correct answee...i m trying to do @mayankdevnani method

Answer 31

@mayankdevnani can u do this question by my method...??

Answer 32

What is your method ?

Answer 33

I described it already ...just scroll it upwards

Answer 34

Jaipur....

Answer 35

Wow! itna paas

Answer 36

Haan chlo ab ques ka b btao...

Answer 37

Dekh liya kya upr krke...

Answer 38

solve kr raha hoon thoda wait kro

Answer 39

Acccha kro fir toh...

Answer 40

you can't solve

Answer 41

Ans match ni kr ra na

Answer 42

integral main t aur x dono function aa rahein h

Answer 43

isliye solve nahi hoga

Answer 44

@ganeshie8 ne galat likh diya integral

Answer 45

Arc sint is d integral of it....i solved it myself also n we have to put t =sinx-cosx afterwards..

Answer 46

okay
now when you replace dx by dt,you have both variables

Answer 47

but ganeshie forgot to wrote that

Answer 48

How come u r saying that v have both variables in just one equation..... I never came across this condition in my method....

Answer 49

okay,write dt first

Answer 50

dt =(cosx + sinx)dx

Answer 51

Vich is the same as the numerator of the question

Answer 52

now denominator becomes 1-t^2

Answer 53

Under radical sign

Answer 54

now understand !

Answer 55

i mean we can't write numerator in terms of `t`

Answer 56

ok...let's leave it....the method of ganeshie 8 i m talking about my way..ok..

Answer 57

@samigupta8 try to understand

Answer 58

Tanx n cot x can b opened up in the form of sin n cos functions

Answer 59

you said \[\large \bf \sqrt{1-t^2}=\sqrt{2sinxcosx}\]
and you are using this

Answer 60

that is mistake !

Answer 61

Just a √2 will come in the numerator... Now is it okay

Answer 62

so we get
\[\large \bf \sqrt{2}\sin^{-1}({sinx-cosx})\]

Answer 63

exactly

Answer 64

bt the ans says there is a negative sign too

Answer 65

This is a failed attempt, but thought I'd share. I've tried to find some kind of general thing for integrals of this form where \(f(x) = \sqrt{\tan x}\)

\[\int f(x) + \frac{1}{f(x)} dx\]
by eventually trying to work up to this substitution, g(f(x)) starting from here to see if I can wiggle something out:
\[g(x) = x+\frac{1}{x}\]

Since I realized:
\[g(x) = g(\tfrac{1}{x}) \]

I could make this substitution in an integral:

\[\int g(x) dx = \int g(\tfrac{1}{x}) dx\]
Or alternatively,
\[x=\tfrac{1}{t}\]
\[dx= \tfrac{-1}{t^2} dt\]
\[\int g(x) dx = \int g(\tfrac{1}{t}) (\tfrac{-1}{t^2}) dt\]

In other words for bounds which are 1/x of each other such as from 2/3 to 3/2 this integral is true:

\[\int g(x) dx = \int -\frac{g(x)}{x^2} dx\]

Anyways I think this ended up not getting us anywhere useful but it got me somewhere interesting, so thanks for the question.

Answer 66

Great Work ! @Kainui

Answer 67

But it seems useless !
I mean for this question

Answer 68

@kainui how can u arrive at such a method ....?

Answer 69

I say from the start it's a failed method, I'm just looking to brainstorm here to see what there is down this line of thinking for fun since this question pretty much wrapped up.

Answer 70

I GOT IT WHERE -VE SIGN CAME

Answer 71

pls..tell now.....how did u get it?

Answer 72

AS x LIES BETWEEN THIRD QUADRANT,
SO TANX AND COT X SHOULD BE POSITIVE

Answer 73

yup.....

Answer 74

BUT OUR ANSWER IS NEGATIVE BECOME SINX AND COSX IN THIRD QUADRANT IS NEGATIVE

Answer 75

SO WE ADD -VE SIGN TO IT

Answer 76

FOR BECOME POSITIVE

Answer 77

understood ?

Answer 78

tanx n citx have to be always greater than 0 in order that no negative sign comes under radical

Answer 79

cotx*

Answer 80

N btw...y do u want that the integration shud be positive??

Answer 81

I think there are two answers...lol

Answer 82

both +ve and -ve

Answer 83

Lol.....two answers for the same question it's not a multicorrect type question ....

Answer 84

Put x=210 degrees, +ve=+ve
When we add -ve sign to it,
Put x=240 degrees, +ve=+ve

Answer 85

^^My OBSERVATION^^

Answer 86

@Kainui