IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

mona840

(cot ^-1 ( e^x) )/ ( e^x) dx

options :-

(1) 1/2 ln ( e^2x+ 1) - (cot ^-1 ( e^x) )/ ( e^x) + x+ C

(2) 1/2 ln ( e^2x+ 1) + (cot ^-1 ( e^x) )/ ( e^x) + x+ C

(3) 1/2 ln ( e^2x+ 1) - (cot ^-1 ( e^x) )/ ( e^x) - x+ C

(4) 1/2 ln ( e^2x+ 1) + (cot ^-1 ( e^x) )/ ( e^x) - x+ C

shubham_sachdeva

DIFFERENTIATE THE OPTIONS ..... WATEVER OPTION THAT WILL GET THE QN. WILL BE THE RIGHT ANSWER......

HOPE U GET IT><

mona840

ya

but hw do v solve it !!!!!!!

aditi_g

see its very simple....just multiply and divide the given expression by e^x and thn put e^x as t.... the expression becomes... cot ^ -1 t/t^2 dt

thn apply that by parts formula of integration and u can easily get the answer... option 3 is correct....and 1 piece of advice at the first attempt nvr try to solve any integration prob by differentiating the options... its just a waste of time... u can use it in case ur sitting for an examination and not getting any answer

pradnya

put e^x=t
x=logt
dx=dt/ t
int becomes--cot^-1t/ t^2 dt

shubham_sachdeva

when u hv options den why to do it properly...sum qns. are meant to be done by done by these methods..

astrojith

Shubham, its not easy to differentiate these. And what if the answer is the 4th one ? You'll lose a lot of time.

And yo mona, is that you in the picture ?

richa_dpsvk

yeah...differentiating every expression would be a real waste of time!!

aditya_arora04

Hi,

Look,

(cot ^-1 ( e^x) )/ ( e^x) dx

(cot ^-1 ( e^x) )*e^x/ ( e^2x) dx ------------------ 1

Put t = e^x ------------------- 2

dt = e^x dx --------------------- 3

cot ^-1 (t) / ( t²) dx

Now, apply product rule by taking cot^-1(t) as first function and t² as second function.

After that you should be able to solve this ques.

Hope u got it !!!!!!

Do tell me in case of problems !!!!

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