IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

nandu

FIND(with detailed proof,if possible):

1.

dx/1+x⁴.

2.

(sinx)^1/2dx.

3.

e^adx.........where,a=x^2.

vasanth

http://integrals.wolfram.com/index.jsp

refer to this......created by wolfram......it 's a part of mathematica.....

1) involves a lot of inverse properties

2)non integrable.......this Q has already bin asked...polylogarithmic fn...beyond scope of JEE

3)beyond JEE scope........involves erfi[] functions.......

check out the link......

plzzz rate me if u found the link useful

cheeeeers

aman23iit

hi friend the solution is

dx/1+x⁴.=int.1/2[(1+x^2)/1+x^4dx]-int.[(1-x^2)/1+x^4dx]

from above take x^2 common in the first case you will see that

(1/x^2+1)/(x^2+1/x^2-2+2)

(1/x^2+1)/[(x-1/x)^2+(2)^2]

put x-1/x=t and dt=dx(1/x^2+1)

and you can solve further eazily in next case also

2. (sinx)^1/2dx is a disscontinuous function and hence cannot be integrated without giving limits

3.

e^adx.........where,a=x^2.e^x=t then you will see that

e^xe^xdx= tdt=t²/2=e^a/2

bye

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bindhugadulocal

in 1st qtn can we take x^2 as t

nishant

the question will become too lenghty then..the solution of aman is the accurate soln..

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