IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

aman531

cn u plz solvethis integration question:

sin x dx

i know its long

i requirre basic steps

plz help

bujji

Put Sinx=t2. the part inside the integral becomes

t2/1-t4.

Apply partial fractions.

KAB

Hey aman531,
We cannot evaluate it.Suppose x is in 3rd or 4th quadrant then root(sinx) will be a complex number.But a definite integral like limit from 0 to pi/2 etc can be evaluated.

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

bujji

If the question is as u have given then it implies that Sinx is positive.

caprio.nups

KAB is rite..this function is non-integrable...

bujji

If Sinx is replaced by x ,does the question give whether x>0 or x<0.

kvsp

very true.. the fn is integrable.. i dont think one has to take into consideration the sign..

KAB

Why dont u guys check this http://integrals.wolfram.com/index.jsp

Just type the function there and u will get it integrated.......Simply dont waste time on such questions.

aman531

rt kab

and thank u all

ginny02

hey u all guys
i had solved it
solve it by parts by taking 1 as 2nd func
so what u all think

devesh_l2k007

-tan(x) + 1/2 log 1+x
1-x

x=root(sin x) tanx = tan (inverse(x) )

devesh_l2k007

put sinx = t^2 cosxdx=2tdt
then cosx=1-t^4

hence exp becomes

2tdt*t
1-t^4

now split as t^2-1 and t^2+1

the job is done then

devesh_l2k007

and abt the link....
see in core Mathematics
thereare many functions whom we are not sure abt but they ar relatively new and are mostly used by mathematecians of the day

so we cant rely upon dem as we don know these kind of things

devesh_l2k007

And to you all y do you think that complex nos cant be used in calculus

They can like vectors

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