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KEAM 2015 Math Paper
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© examsnet.com
Question : 6
Total: 120
∫
(
1
+
x
)
e
x
c
o
t
(
x
e
x
)
d
x
is equal to
log
|
c
o
s
(
x
e
x
)
|
+
C
log
|
c
o
t
(
x
e
x
)
|
+
C
log
|
s
e
c
(
x
e
−
x
)
|
+
C
log
|
c
o
s
(
x
e
−
x
)
|
+
C
log
|
s
e
c
(
x
e
x
)
|
+
C
Validate
Solution:
Let
I
=
∫
(
1
+
x
)
e
x
cot
(
x
e
x
)
d
x
Put
x
e
x
=
t
⇒
(
x
e
x
+
e
x
)
d
x
=
d
t
⇒
(
x
+
1
)
e
x
d
x
=
d
t
∴
I
=
∫
d
t
cot
t
=
∫
tan
t
d
t
=
log
|
sec
t
|
+
C
Again, put
t
=
x
e
x
⇒
I
=
log
|
sec
(
x
e
x
)
|
+
C
© examsnet.com
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