NDA Math Indefinite Integrals

Let us differentiate all the options one by one to get the expression in the question whose integral is to be found.
Here xesin‌x is the common term in all the options. So,let us differentiate it first.
Let l=xesin‌x
⇒‌‌

dl

dx

=esinx[x‌cosx+1]
⇒‌‌

dl

dx

=‌‌

esinx

cos2x

[xcos3x+cos2x]
Let m=sec‌xesin‌x
⇒‌‌

dm

dx

=esinx[1+‌‌

sinx

cos2x

]
⇒‌‌

dm

dx

=‌‌

esinx

cos2x

[cos2x+sinx]
Differentiation of option (a) is
=‌‌

esinx

cos2x

[xcos3x+2cos2x+sinx]
Differentiation of option (b) is
=‌‌

esinx

cos2x

[xcos3x−sinx]
∴ Option (b) is correct.