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Byju's Answer
Standard XII
Physics
Antiderivative
∫ tan x+2tan2...
Question
∫
(
tan
x
+
2
tan
2
x
+
4
tan
4
x
+
8
cot
8
x
)
d
x
has the value
A
ln
(
sin
x
)
+
c
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B
−
ln
(
cosec
x
)
+
c
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C
ln
(
cos
x
)
+
c
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D
ln
(
sin
2
x
)
−
ln
(
2
cos
x
)
+
c
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Solution
The correct options are
A
ln
(
sin
x
)
+
c
B
−
ln
(
cosec
x
)
+
c
Let
I
=
∫
(
tan
x
+
2
tan
2
x
+
4
tan
4
x
+
8
cot
8
x
)
d
x
tan
x
−
cot
x
=
tan
x
−
1
tan
x
=
(
tan
2
x
−
1
tan
x
)
×
2
×
2
=
2
−
tan
2
x
=
−
2
cot
2
x
Back to integral, adding and subtracting
cot
x
, we get
I
=
∫
(
tan
x
−
cot
x
+
2
tan
2
x
+
4
tan
4
x
+
8
cot
8
x
+
cot
x
)
d
x
=
∫
(
−
2
cot
2
x
+
2
tan
2
x
+
4
tan
4
x
+
8
cot
8
x
+
cot
x
)
d
x
=
∫
cot
x
Now, we have to integrate
cot
x
I
=
∫
cot
x
=
ln
sin
x
+
c
=
−
ln
(
cosec
x
)
+
c
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0
Similar questions
Q.
If
tan
x
+
2
tan
2
x
+
4
tan
4
x
+
8
cot
8
x
=
√
3
then the general solution of
x
=
Q.
Differentiate
l
o
g
|
tan
x
|
+
1
2
tan
2
x
+
C
.
Q.
∫
d
x
s
i
n
4
x
+
c
o
s
4
x
is equal to
Q.
If
∫
tan
x
sec
x
+
tan
x
d
x
=
I
+
x
+
c
where
c
is an arbitary constant, then the value of
I
is
Q.
If
∫
sin
x
sin
3
x
+
cos
3
x
d
x
=
α
log
e
|
1
+
tan
x
|
+
β
log
e
|
1
−
tan
x
+
tan
2
x
|
+
γ
tan
−
1
(
2
tan
x
−
1
√
3
)
+
C
,
where
C
is constant of integration, then the value of
18
(
α
+
β
+
γ
2
)
is
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