Question

integrate [ sqrt (tanx) + sqrt (cotx) ] dx

limits of integration:
lower limit = 0
upper limit = pi/4

prove: it'll be equal to
sqrt(2) . (pi /2)

need hints. cant get any idea :(

Answer 1

As i answered yesterday one of the questions, you need to think of conversion here, in other words thinking of this trig function in terms of 1/(1+x^4)

Answer 2

you need to use u substitution

Answer 3

this is the answer for integral of sqrt

(1/4)*sqrt(2)*(2*sqrt(sin(x)/cos(x))*cos(x)*arccos(cos(x)-sin(x))+sqrt(cos(x)*sin(x))*ln(2)-2*sqrt(cos(x)*sin(x))*ln(cos(x)*sqrt(2)+2*sqrt(sin(x)/cos(x))*cos(x)+sqrt(2)*sin(x)))/sqrt(cos(x)*sin(x))

Answer 4

i tried using properties of definate integrals... as i saw pi/4 in limits... and tan (pi/4) = 1 = cot (pi/4) but failed.. tried to convert it in sin and cos stuck again..

Answer 5

\[\int\limits \sqrt{\tan(x)}dx\]

Answer 6

i would recommend that you expand the taylor series of each function

Answer 7

no Taylor series for sqrt tan(x), we need to search for different expansion method of any series...

Answer 8

is your teacher mad with his wife or something, to fulfill his anger with such an integral...!!????

Answer 9

i will have to use the whole BW of openstudy to type the solution for this problem...

Answer 10

there would be some easy solution...

Answer 11

let u = tanx, du = sec^xdx, do the same thing for cot(x)

Answer 12

without expantion because we havent bee taught any expantion of tan or any such trignometric functions at school..
i guess properties of integral must simplyfy things..
i get sqrt (2) . (sin x + cos x) / sqrt (sin2x)

Answer 13

now using properties..
sqrt 2 (sin y+ cos y)/ sqrt (cos 2 x)

y = pi/4 - x

Answer 14

now at this step i am stuck.. i dont know how to convert sin x + cos x as a derivative of denominator or sumthing like that..

Answer 15

well even if you use tanx^2=1-csc^2 property, you will gave to take the 4rth root...!!!!

Answer 16

actually let me try a different method...

Answer 17

tanx = sinx/cosx

Answer 18

ohkay . . .

Answer 19

cotx=cosx/sinx

Answer 20

but we will take the root of them

Answer 21

did you try to simplify this expression

Answer 22

\[\int\limits \sqrt{tanx}+\sqrt{cotx}dx =>\int\limits\frac{sinx+cosx}{\sqrt{sinxcosx}} => \]
\[u=sinx-cosx => du=(sinx+cosx)dx\]
\[1-2\sin2x=u^2=> \sin2x=(1-u^2)\]
\[\int\limits\limits\frac{\sin+cosx}{\sqrt{1/2\sin2x}} =>\sqrt{2} \int\limits\limits\frac{sinx+cosx}{\sqrt{\sin2x}} =>\sqrt{2}\int\limits\limits\frac{1}{\sqrt{1-u^{2}}} du\]

Answer 23

^WOW O-O

Answer 24

mathsmind, i wanted to do the first few steps... rest of steps were mystery to me...
thanks! mimi_x3

Answer 25

all the best

Answer 26

MIWIWIWIWIWIWIWIWIWIWI