E-StoreHome » Ask & Discuss » Mathematics. » Integral Calculus « Back to Discussion
Integral Calculus
3 Feb 2012 22:38:26 IST
0 People liked this
4
270
Comments (4)
4 Feb 2012 19:29:07 IST
Like
0 people liked this
Putting 
We get 
The question now becomes
Writing 2t2 as (1+t2) - (1-t2), we get 2 integrals.
Divide numerator and denominator by t2 in both the integrals to get the 2 integrals as
In first integral, put t - 1/t = u
In second integral, put t + 1/t = v
Integrate....hope you can do the rest.........
Free Sign Up!
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
Find Posts by Topics
Physics.
Topics
9395
20091
3672
2186
7617
2827
2637
Mathematics.
Chemistry.
Biology
Institutes
Parents
Board
Fun Zone
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
921 Students Currently Online |
Connect with Us  |
 |
Vriti Communities
Privacy Policy © Vriti 2011
U have to take whole as t...
let (tan x)^1/2 = t
tan x = t^2
sec^2 x dx = 2t dt [sec^2 x = tan^2 x + 1]
dx = 2t / *( t^4 + 1) dt
so,
(int) t^2 * 2 t / ( t^4 + 1) dt
(int) [2 t^3 / ( t^4 + 1) ] dt
2( t^4 + 1) / 3 + c (constant of integration)
2 (tan^2 x + 1) / 3 + c..........!!!