How would you solve this integral problem below? - correct to six decimal places-?

Question

2.3
∫ [(cos x)/x] dx =
1.9

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it has to be one of these solutions

a. - 0.181097

b. -0.139520

c. -0.094765

d. -0.048752

e. 0.164432

f. 0.246669

g. 0.328928

Answer 1

c. -0.09476473214
Did it on Maple. (I would suggest a series expansion for cosx).

Answer 2

Tailor series: cosx =1-x^2/2 +x^4/4! –x^6/6! ++++
(cos x)/x =1/x –x/2 +x^3/4! –x^5/6! ++++

2.3
∫(1/x –x/2 +x^3/4! –x^5/6! ++++)*dx =
1.9

=lnx –(1/2)x^2/2! +(1/4)x^4/4! –(1/6)x^6/6! ++++

+ln(2.3)-ln(1.9) =ln(2.3/1.9) =0.191055237
-(1/2) (2.3^2 -1.9^2) / 2! =-0.42
+(1/4) (2.3^4 -1.9^4) / 4! =0.15575
-(1/6) (2.3^6 -1.9^6) / 6! =-0.023377317
+(1/8) (2.3^8 -1.9^8) / 8! =0.001901272
-(1/10) (2.3^10 -1.9^10) /10! =-9.72648E-05
+(1/12) (2.3^12 -1.9^12) /12! =3.4275E-06
-(1/14) (2.3^14 -1.9^14) /14! =-8.84381E-08; enough!
____________________total=-0.094764734;
I cannot suggest anything niftier.