Therefore,
2
2
dx
x
−a
∫
 =
2
2
2
secθ tanθ θ
sec θ
a
d
a
−a
∫
=
1
secθ θ
d =log secθ + tanθ + C
∫
=
2
1
2
log
1
C
x
x –
a
a
+
+
=
2
2
1
log
log
C
x
x – a
a
+
−
+
=
2
2
log
+ C
x
x – a
+
, where C = C1 – log |a|
(5) Let x = a sinθ Then dx = a cosθ dθ Therefore,
 
2
2
dx
a
−x
∫
 =
2
2
2
θ θ
θ
cos
sin
a
d
a – a
∫
=
θ = θ + C = sin1
C
–
x
d
a
+
∫
(6) Let x = a tanθ Then dx = a sec2θ dθ