Begin2.DVI

(Ben Green) #1




∫ 1
xcsc

− 1 x
adx=−

(
a
x+

1
2 · 3 · 3

(x
a

) 3
+ 2 ·^14 ··^35 · 5

(x
a

) 5
+ 2 ·^14 ·· 63 ··^57 · 7

(x
a

) 7
+···

)
+C





∫ 1
x^2 sin

− 1 x
adx=−

1
xsin

− 1 x
a−

1
aln|

a+


a^2 −x^2
a |+C





∫ 1
x^2 cos

− 1 x
adx=−

1
xcos

− 1 x
a+

1
aln|

a+


a^2 −x^2
a |+C





∫ 1
x^2 tan

− 1 x
adx=−

1
xtan

− 1 x
a−

1
2 aln|

x^2 +a^2
a^2 |+C





∫ 1
x^2 cot

− 1 x
adx=−

1
xcot

− 1 x
a+

1
2 a

∫ 1
xtan

− 1 x
adx





∫ 1
x^2 sec

− 1 x
adx=





−^1 xsec−^1 xa+ax^1


x^2 −a^2 +C, 0 <sec−^1 xa< π/ 2

−^1 xsec−^1 xa−ax^1


x^2 −a^2 +C, π/ 2 <sec−^1 xa< π





∫ 1
x^2 csc

− 1 x
adx=





−^1 xcsc−^1 xa−ax^1


x^2 −a^2 +C, 0 <csc−^1 xa< π/ 2

−^1 xcsc−^1 xa+ax^1


x^2 −a^2 +C, −π/ 2 <csc−^1 xa< 0






sin−^1

√ x
a+xdx= (a+x) tan

− 1


x
a−


ax+C






cos−^1

√ x
a+xdx= (2a+x) tan

− 1

√x
2 a−


2 ax+C

Integrals containing the exponential function






eaxdx=^1 aeax+C






xeaxdx=

(x
a−

1
a^2

)
eax+C






x^2 eaxdx=

(x 2
a −

2 x
a^2 +

2
a^3

)
eax+C






xneaxdx=^1 axneax−na


xn−^1 eaxdx





∫ 1
xe

axdx= ln|x|+ ax
1 ·1!+

(ax)^2
2 ·2!+

(ax)^3
3 ·3!+···+C





∫ 1
xne

axdx=−^1
(n−1)xn−^1 e

ax+ a
n− 1

∫ 1
xn−^1 e

axdx





∫ eax
α+βeaxdx=

1
aβln|α+βe

ax|+C

Appendix C
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