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Byju's Answer
Standard XII
Mathematics
Differentiation Using Substitution
Evaluate the ...
Question
Evaluate the following integrals:
∫
e
2
x
sin
3
x
+
1
d
x
Open in App
Solution
We
have
,
I
=
∫
e
2
x
sin
3
x
+
1
d
x
Let
the
first
function
be
sin
3
x
+
1
and
the
second
function
be
e
2
x
.
First
we
find
the
integral
of
the
second
function
,
i
.
e
.
,
∫
e
2
x
dx
.
∫
e
2
x
dx
=
1
2
e
2
x
Now
,
using
integration
by
parts
,
we
get
I
=
sin
3
x
+
1
∫
e
2
x
d
x
-
∫
d
sin
3
x
+
1
d
x
∫
e
2
x
d
x
d
x
=
1
2
sin
3
x
+
1
e
2
x
-
3
2
∫
cos
3
x
+
1
e
2
x
d
x
=
1
2
sin
3
x
+
1
e
2
x
-
3
2
cos
3
x
+
1
∫
e
2
x
d
x
-
∫
d
cos
3
x
+
1
d
x
∫
e
2
x
d
x
d
x
=
1
2
sin
3
x
+
1
e
2
x
-
3
2
1
2
cos
3
x
+
1
e
2
x
+
3
2
∫
sin
3
x
+
1
e
2
x
d
x
=
1
2
sin
3
x
+
1
e
2
x
-
3
4
cos
3
x
+
1
e
2
x
-
9
4
I
+
c
I
+
9
4
I
=
1
2
sin
3
x
+
1
e
2
x
-
3
4
cos
3
x
+
1
e
2
x
+
c
13
4
I
=
e
2
x
2
sin
3
x
+
1
-
3
2
cos
3
x
+
1
+
c
I
=
2
13
e
2
x
sin
3
x
+
1
-
3
2
cos
3
x
+
1
+
c
=
e
2
x
13
2
sin
3
x
+
1
-
3
cos
3
x
+
1
+
c
Hence
,
∫
e
2
x
sin
3
x
+
1
d
x
=
e
2
x
13
2
sin
3
x
+
1
-
3
cos
3
x
+
1
+
c
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