Begin2.DVI

∫

eaxsinbx dx=

(

asinbx−bcosbx

a^2 +b^2

)

eax+C

∫

eaxcosbx dx=

(

acosbx+bsinbx

a^2 +b^2

)

eax+C

∫

eaxsinnbx dx=

(asinbx−nbcosbx

a^2 +n^2 b^2

)

eaxsinn−^1 bx+n(n−1)b

2

a^2 +n^2 b^2

∫

eaxsinn−^2 bx dx

∫

eaxcosnbx dx=

(acosbx+nbsinbx

a^2 +n^2 b^2

)

eaxcosn−^1 bx+n(n−1)b

2

a^2 +n^2 b^2

∫

eaxcosn−^2 bx dx

Another way to express the above integrals is to define

Cn=

∫

eaxcosnbx dxandSn=

∫

eaxsinnbx dx, then one can write the reduction formulas

Cn=acosabx (^2) ++nnb (^2) b 2 sinbxeaxcosn−^1 bx+n(n−1)b
2
a^2 +n^2 b^2 Cn−^2
Sn=asinabx (^2) +−nnb (^2) bcos 2 bxeaxsinn−^1 bx+n(n−1)b
2
a^2 +n^2 b^2 Sn−^2
558.
∫
xeaxsinbx dx=
([2ab−b(a (^2) +b (^2) )x] cosbx+ [a(a (^2) +b (^2) )x−a (^2) +b (^2) ] sinbx
(a^2 +b^2 )^2
)
eax+C
559.
∫
xeaxcosbx dx=
([a(a (^2) +b (^2) )x−a (^2) +b (^2) ] cosbx+ [b(a (^2) +b (^2) )x− 2 ab] sinbx
(a^2 +b^2 )^2
)
eax+C
560.
∫
eaxlnx dx=a^1 eaxlnx−a^1
∫ 1
xe
axdx
561.
∫
eaxsinhbx dx=
[asinhbx−bcoshbx
(a−b)(a+b)
]
eax+C, a 6 =b
562.
∫
eaxsinhax dx= 41 ae^2 ax−x 2 +C
563.
∫
eaxcoshbx dx=
[acoshbx−bsinhbx
(a−b)(a+b)
]
eax+C, a 6 =b
564.
∫
eaxcoshax dx= 41 ae^2 ax+x 2 +C
565.
∫ dx
α+βeax=
x
α−
1
aαln|α+βe
ax|+C
566.
∫ dx
(α+βeax)^2 =
x
α^2 +
1
aα(α+βeax)−
1
aα^2 ln|α+βe
ax|+C
567.
∫ dx
αeax+βe−ax=





1
a
√
αβ
tan−^1
(√α
βe
ax
)
+C, αβ > 0
1
2 a
√
−αβln
∣∣
∣∣
∣
eax−
√
−β/α
eax+
√
−β/α
∣∣
∣∣
∣+C, αβ <^0
Appendix C