[x ]
[0 ] [0 ]
[x ] t2/ (x-sinx)
a+t dt
As the integral extends only over the variable t, hence it can be written as
[ x][0 ] [ [0 ][x ] (t2 /a+t) dt] /(x-sinx)
Now, since it is 0/0 form, hence apply L'Hospital's Rule, and in the numerator, Leibnitz rule will be applied.
[x ][0 ] (x2/a+x) / (1-cosx)
[x ][ 0] 2/ a+x (sin2(x/2) /(x2/4) )
on applying the limit, u get
2/a =1
Hence, a=4
Moderation Team
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