IIT JEE , AIEEE Forum - Mathematics , Algebra - does anyone dare 2 solve this??????????!!!!!!!!!!!!!!!!!!!!!!!!!!!!

ank_einstein

evaluate::::::::::::::::

1)

sqrt(cos2x) /sinx dx...........................

2) if I_n =

(sinx)ⁿ/cos²2x dx .............

then I1 ,I2,I3........... are in

(A) AP (b) GP (C) HP (D) NONE OF THESE...................

elessar_iitkgp

Transform the integral as follows

I =

(

cos2x)/sinx =

(

(cos² x - sin² x))/sinx =

(

(1 - tan² x))/tanx
(Dividing the Nr & Dr by cosx)

Substitute tanx = t

I =

[(

(1-t² ))/t][1/(1+t² )] dt =

[(

(1-t² ))/t(1+t² )] dt

Substitute 1-t² = u²

I = -

[u /t(2-u² )][u/t} du = -

[u² /(1-u² )(2-u² )]du =

[u² /(u² -1)(2-u² )]du
=(1/2)

[2(u² -1) + (2-u² )/(u² -1)(2-u² )]du = (1/2)[

2/( (2-u² ) du -

1/(1-u²) du]
= (1/

2)ln[(

2 + u)/(

2 - u)] - (1/ 2)ln[(

2 + u)/(

2 - u)]
=(1/

2)ln[(

2 +

(1 - tan² x))/(

2 -

(1 - tan² x))] - (1/ 2)ln[(

2 +

(1 - tan² x))/(

2 -

(1 - tan² x))]
=(1/

2)ln[(

2 cosx +

(cos² x - sin² x))/(

2 cosx -

(cos² x - sin² x))] - (1/ 2)ln[(

2 cosx +

(cos² x - sin² x))/(

2 cosx -

(cos² x - sin² x))]
= (1/

2)ln[(

2 cosx +

(cos2x))/(

2 cosx -

(cos2x)]
- (1/ 2)ln[(

2 cosx +

(cos2x))/(

2 cosx -

(cos2x))]

Hope the answer si right ... Lengthy ...phew!!

elessar_iitkgp

Plz make sure u didn't forget to include the limits of integration in Q2

a_buddy4uin06

Q1 in a bit easier way.... if ter is any mistake in this pla point out... cos i m not tat gr8 in integration.. so here goes..

(cos²x - sin²x)/ sinx dx
=

(1 - 2sin²x)/

sin²x dx
=

(1/ sin²x - 1)dx
=

(cosec²x - 1)dx
=

cotx dx
=log sinx + c

a_buddy4uin06

i believe d sq. root is ter only fr numerator..

elessar_iitkgp

Check the third step. There should be a 2 there

Ask & Discuss Questions with Community & Experts