Then dx = a sec2 θ dθ Therefore,      
2
2
dx
x
+a
∫
 = 
2
2
2
2
θ
θ
secθ
tan
a
d
a
+a
∫
=
1
1
1
1
θ
θ
C
tan
C
– x
d
a
a
a
a
=
+
=
+
∫
(4) Let x = a secθ Then dx = a secθ tan θ d θ 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