Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

ln(sec x+yan x)/cos x dx=ln(sec y+tan y)/cosy dy

ln(sec x+yan x)/cos x dx=ln(sec y+tan y)/cosy dy

Grade:12

1 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear upendra

ln(sec x+yan x)/cos x dx=ln(sec y+tan y)/cosy dy

integrate both side

ln(sec x+yan x)/cos x dx=ln(sec y+tan y)/cosy dy

 

let secx +tan x =u                               and sexy +tan y =t

  (secx tanx +sec2x ) dx =du

or secx (secx +tanx) dx=du

or dx/cosx  =du/u

 

so intigral become

 1/u logu   du  =1/t logt   dt

let I=1/u logu   du

intigrate using first and second finction

I=logu*logu -1/u logu   du

2I=(logu)2

I=1/2 (logu)2


similerly for right hand side

intigral become

1/2 (logu)2     = 1/2 (logt)2     +C

 (log(secx+tanx))2     (log(secy+tany))2     +C


Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin


Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free