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Byju's Answer
Standard XII
Mathematics
Properties of Determinants
If = |[ 2...
Question
If
△
=
∣
∣ ∣ ∣
∣
2
c
o
s
2
x
s
i
n
2
x
−
s
i
n
x
s
i
n
2
x
2
s
i
n
2
x
c
o
s
x
s
i
n
x
−
c
o
s
x
0
∣
∣ ∣ ∣
∣
Prove that
∫
π
/
2
0
[
△
+
△
′
]
d
x
=
π
where
△
′
=
d
/
d
x
(
△
)
.
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Solution
Apply
C
1
−
2
s
i
n
x
C
3
,
C
2
+
2
c
o
s
X
C
3
△
=
∣
∣ ∣
∣
2
0
−
s
i
n
x
0
2
c
o
s
x
s
i
n
x
−
c
o
s
x
0
∣
∣ ∣
∣
=
2
(
c
o
s
2
x
+
s
i
n
x
)
=
2
∴
△
′
=
0
a
n
d
△
+
△
′
=
2
.
∴
∫
π
/
2
0
(
△
+
△
′
)
d
x
=
∫
π
/
2
0
2
d
x
=
[
2
x
]
π
/
2
0
=
π
.
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0
Similar questions
Q.
∫
π
/
2
0
(
|
sin
x
|
+
|
cos
x
|
)
d
x
Q.
Let
∣
∣ ∣ ∣
∣
2
cos
2
x
sin
(
2
x
)
−
sin
x
sin
(
2
x
)
2
sin
2
x
cos
x
sin
x
−
cos
x
0
∣
∣ ∣ ∣
∣
Find
1
π
∫
π
/
2
0
[
f
(
x
)
+
f
′
(
x
)
]
d
x
Q.
If
y
=
cot
−
1
√
1
+
sin
x
+
√
1
−
sin
x
√
1
+
sin
x
−
√
1
−
sin
x
, find
d
y
d
x
if
x
∈
(
0
,
π
2
)
∪
(
π
2
,
π
)
Q.
If
y
=
cot
−
1
[
√
1
+
sin
x
+
√
1
−
sin
x
√
1
+
sin
x
−
√
1
−
sin
x
]
, where
0
<
x
<
π
2
, then
d
y
d
x
is equal to
Q.
Solve:
π
/
2
∫
0
|
sin
x
−
cos
x
|
d
x
=
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