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Question
Mathematics
For n ≥ 2, If In=∫ sec n x d x, then I4-(2/3) I2=
Q. For
n
≥
2
, If
I
n
=
∫
sec
n
x
d
x
, then
I
4
−
3
2
I
2
=
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A
sec
2
x
tan
x
+
c
B
3
1
sec
2
x
tan
x
+
c
C
3
2
sec
2
x
tan
x
+
c
D
2
1
lo
g
∣
sec
x
+
tan
x
∣
+
c
Solution:
We have,
I
n
=
∫
sec
n
x
d
x
∴
I
2
=
∫
sec
2
x
d
x
=
tan
x
+
c
1
and
I
4
=
∫
sec
4
x
d
x
=
∫
sec
2
x
⋅
sec
2
x
d
x
=
∫
(
tan
2
x
+
1
)
sec
2
x
d
x
=
3
t
a
n
3
x
+
tan
x
+
c
2
∴
I
4
−
3
2
I
2
=
3
t
a
n
3
x
+
tan
x
+
c
2
−
3
2
tan
x
−
3
2
c
1
=
3
1
tan
3
x
+
3
1
tan
x
+
c
[
where
c
2
−
3
2
c
1
=
c
]
=
3
1
tan
x
(
tan
2
x
+
1
)
+
c
=
3
1
tan
x
sec
2
x
+
c