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Byju's Answer
Standard XII
Mathematics
Integration of Trigonometric Functions
Solve ∫ x √...
Question
Solve
∫
cot
x
√
cos
4
x
+
sin
4
x
d
x
=
.
.
.
.
.
.
+
c
A
1
2
log
∣
∣
cot
2
x
+
√
cot
4
x
+
1
∣
∣
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B
−
1
2
log
∣
∣
cot
2
x
+
√
cot
4
x
+
1
∣
∣
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C
1
2
log
∣
∣
tan
2
x
+
√
tan
4
x
+
1
∣
∣
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D
−
1
2
log
∣
∣
cot
x
+
√
cot
4
x
+
1
∣
∣
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Solution
The correct option is
B
−
1
2
log
∣
∣
cot
2
x
+
√
cot
4
x
+
1
∣
∣
I
=
∫
cot
x
√
cos
4
x
+
sin
4
x
d
x
I
=
∫
cot
x
⎷
sin
4
x
(
cos
4
x
sin
4
x
+
1
)
d
x
I
=
∫
cot
x
sin
2
x
√
cot
4
x
+
1
d
x
I
=
∫
cot
x
c
o
sec
2
x
√
cot
4
x
+
1
d
x
Now take
cot
2
x
=
t
∴
−
2
cot
x
c
o
sec
2
x
d
x
=
d
t
∴
cot
x
c
o
sec
2
x
d
x
=
−
1
2
d
t
∴
I
=
−
1
2
∫
1
√
t
2
+
1
d
t
=
−
1
2
log
∣
∣
t
+
√
t
2
+
1
∣
∣
+
c
∴
I
=
−
1
2
log
∣
∣
cot
2
x
+
√
cot
4
x
+
1
∣
∣
+
c
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is equal to