seems easy bt nt getting

I1=∫(x/sinx)

int frm 0 to pi/2

I2=∫arc tanx/x
int frm 0 to 1

then I1/I2 =?

8 Answers

1357
Manish Shankar ·

I2=∫arc tanx/x
int frm 0 to 1

put arctanx=t

I2=(0 to π/4)∫t.sec2tdt/tant=∫t/sintcost=∫2tdt/sin2t

put 2t=x it becomes (0 to pi/2)∫x/sinx

1
The Race begins... ·

a small editing bro.!

put tan-1x (or arc tanx)=t. and the rest of the solution is same.

1
skygirl ·

wat is arc tanx ?

1357
Manish Shankar ·

hmm

editing my post

1
skygirl ·

1.) I= 0∫Π/2 x/sinx dx

=> I= 1/2 0∫Πx/sinx dx -------#1

=> I=1/2 0∫ΠΠ-x/sinx dx ----------#2

adding #1` and #2,

2I = 0∫ΠΠdx => I =Π2/2.

1357
Manish Shankar ·

sky

it is tan-1x

1
°ღ•๓яυΠ·

oh thanks bhayah i wantd d same

1
skygirl ·

okie :)

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