def

∫[tan^(-1)x]^2dx limit is 0 to 1

42 Answers

1
ankit mahapatra ·

nahi hua

13
Двҥїяuρ now in medical c ·

hey aman how did u integrate t*tan t

1
MATRIX ·

yaa aman hw did u integrate after the philip'sssss method........[7][7][7]

1
big looser ......... ·

arey abhirup mene b to wo hi pucha hai

13
Двҥїяuρ now in medical c ·

[7][7]

1
MATRIX ·

ok....

1
MATRIX ·

[9][9][9][9][9][9][7][7][7][7][7][11][11][11][11]

1
big looser ......... ·

hua ya nai ???????????????????

1
MATRIX ·

nahiiiiiiiii..........[7][7][7]

13
Двҥїяuρ now in medical c ·

indefinite does not exist...i m sure.....dunno about definite[7]

1
MATRIX ·

bullgaya kya???.....definite kehh baree mehhh...abhi.......[1][1][1]

1
big looser ......... ·

lekin philip ne direct hi integrate kar dia. limit will be 0,Ï€/4

62
Lokesh Verma ·

nahi hua :(

11
Mani Pal Singh ·

HOW COULD THIS QUESTION BE LEFT UNSOLVED[1]

1
Philip Calvert ·

im not sure mani.......

assumin tan inverse never goes negative in given interval......

integral is 0.245281203139486 i dont know whther ur expression also gives same value

11
Mani Pal Singh ·

yaar check kar le agaar theek hai to badia
aggar galat hai to phir se kar doonga[1]

39
Dr.House ·

fourier series and catalan's constant

11
Mani Pal Singh ·

haan haan wohi

par woh hai kya[3]

1
KR ·

http://targetiit.com/iit_jee_forum/posts/aisa_hota_hai_kya_3317.html

#12

11
Mani Pal Singh ·

KR bhai this ques is different

find the difference (spot the not)[1]

39
Dr.House ·

wrt post no 39, see this

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1072333#1072333

there u see solution by carcul.

1
Philip Calvert ·

ok what i meant was

by using by parts straight away we get

x(tan-1x)2 - ∫ (2xtan-1x/1+x2)dx

tan-1x = t

==> ∫2t.tant dt

note that second term by both methods is same [11]
i wonder whether limits will be same as well
wow mathemagic !![4]

1
big looser ......... ·

ye tan inverse x ka whole square hai

1
Philip Calvert ·

no aman she means dx
is not there and btw sky wrong post urs it shud be ?? not //[3]

1
big looser ......... ·

sorry friends now i have corrected

1
Philip Calvert ·

take (tan^(-1)x)^2 as first funcion and 1 as 2nd and use by parts method

for the second term use again by parts
dont' forget to simplify the second term i think it will be ∫tan^-1 t dt

sorry it will be a dirty method but should give answer

1
prateek punj ·

hey guys cann't we solve it by taking tan-1x=t.....

just a guess....

1
skygirl ·

yeah... take as prateek tld...

1
Philip Calvert ·

by putting tan-1x=t and assuming i am not sleeping(which i am)
it will be
∫t2/sec2t dt

what then [7][7]

*edited sorry it will be ∫t2sec2t dt
anyways what next [7]

1
prateek punj ·

@philip... it is by ur method or mine....

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