(1)
Taking ex
 and cos x as the first and second functions, respectively, in I1, we get
I1 =
sin
sin
x
x
e
x – e
x dx
∫
Substituting the value of I1 in (1), we get
I = – ex cos x + ex sin x – I  or  2I = ex (sin x – cos x)
Hence,
I =
sin
(sin
cos ) + C
2
x
x
e
e
x dx
x –
x
=
∫
Alternatively, above integral can also be determined by taking sin x as the first function
and ex the second function 7 6 1 Integral of the type 
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
We have
I =
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
 = 
( )
+
( )
x
x
e f x dx
e f
′x dx
∫
∫
=
1
1
I
( )
, where I =
( )
x
x
e f
x dx
e f x dx
′
+ ∫
∫