Then dx = dt Therefore
sin
sin (
)
sin (
)
sin
x
t – a
dx
dt
x
a
t
=
+
∫
∫
= 
sin cos
cos sin
sin
t
a –
t
a dt
t
∫
= cos
– sin
cot
a dt
a
t dt
∫
∫
= 
1
(cos )
(sin ) log sin
C
a t –
a
t


+


= 
1
(cos ) (
)
(sin ) log sin (
)
C
a
x
a –
a
x
a


+
+
+


= 
1
cos
cos
(sin ) log sin (
)
C sin
x
a
a
a –
a
x
a –
a
+
+
Hence, 
sin
sin (
)
x
dx
x
+a
∫
 = x cos a – sin a log |sin (x + a)| + C,
where,  C = – C1 sin a + a cos a, is another arbitrary constant (iii)
cos
1
tan
cos
sin
dx
x dx
x
x
x
=
+
+
∫
∫
= 
1
(cos
+ sin
+ cos
– sin )
2
cos
sin
x
x
x
x dx
x
x
+
∫
304
MATHEMATICS
= 
1
1
cos
– sin
2
2
cos
sin
x
x
dx
dx
x
x
+
+
∫
∫
= 
C1
1
cos
sin
2
2
2
cos
sin
x
x –
x dx
x
x
+
+
+
∫ (1)
Now, consider 
cos
sin
I
cos
sin
x –
x dx
x
x
=
+
∫
Put cos x + sin x = t so that (cos x – sin x) dx = dt
Therefore       
2
I
log
C
dt
t
t
=
=
+
∫
= 
2
log cos
sin
C
x
x
+
+
Putting it in (1), we get
1
2
C
C
1
+
+
log cos
sin
1
tan
2
2
2
2
dx
x
x
x
x
=
+
+
+
∫
= 
1
2
C
C
+1
log cos
sin
2
2
2
2
x
x
x
+
+
+
= 
1
2
C
C
+1
log cos
sin
C
C
2
2
2
2
x
x
x
,

+
+
=
+




EXERCISE 7