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Byju's Answer
Standard XII
Mathematics
Logarithmic Differentiation
If ∫0π/2xns...
Question
If
∫
π
2
0
x
n
sin
x
d
x
=
(
3
4
)
(
π
2
−
8
)
. Find n
Open in App
Solution
Consider
L
.
H
.
S
I
n
=
∫
π
2
0
x
n
sin
x
d
x
Evaluating by integration by parts we get
I
n
=
x
n
∫
π
2
0
sin
x
d
x
−
∫
(
∫
π
2
0
sin
x
d
x
)
(
d
d
x
x
n
)
d
x
I
n
=
[
−
x
n
cos
x
]
π
2
0
+
∫
π
2
0
n
x
n
−
1
cos
x
d
x
I
n
=
−
[
(
π
2
)
n
cos
π
2
−
0
n
cos
0
]
+
n
∫
π
2
0
x
n
−
1
cos
x
d
x
I
n
=
n
∫
π
2
0
x
n
−
1
cos
x
d
x
I
n
=
n
[
x
n
−
1
∫
cos
x
d
x
−
∫
(
∫
n
cos
x
d
x
d
d
x
(
x
n
−
1
)
)
d
x
]
I
n
=
n
[
(
x
n
−
1
sin
x
)
π
2
0
−
n
(
n
−
1
)
∫
π
2
0
x
n
−
2
sin
x
d
x
]
I
n
=
n
[
(
π
2
)
n
−
1
sin
π
2
−
0
]
−
(
n
−
1
)
n
∫
π
2
0
x
n
−
2
sin
x
d
x
I
n
=
n
[
π
2
]
n
−
1
−
n
(
n
−
1
)
I
n
−
2
Now put
n
=
3
I
3
=
3
(
π
2
)
2
−
6
I
1
=
3
π
2
4
−
6
I
1
Now to find
I
1
I
1
=
∫
π
2
0
x
sin
x
d
x
=
x
∫
π
2
0
sin
x
d
x
−
∫
π
2
0
sin
x
d
x
=
x
cos
x
π
2
0
+
sin
x
π
2
0
=
(
π
2
cos
π
2
+
0.
cos
0
)
+
(
sin
π
2
−
sin
0
)
=
1
I
3
=
3
π
2
4
−
6
=
3
4
(
π
2
−
24
3
)
=
3
4
(
π
2
−
8
)
So comparing this result with out given equation we can say that
n
=
3
Suggest Corrections
0
Similar questions
Q.
Find
∫
π
2
0
sin
4
x
d
x
if
I
(
n
)
=
∫
π
2
0
sin
n
x
d
x
=
(
n
−
1
)
n
(
n
−
3
)
(
n
−
2
)
(
n
−
5
)
(
n
−
4
)
…
I
0
Q.
Assertion :The value of
∫
π
/
2
0
sin
6
x
d
x
=
5
π
16
Reason: If n is even, then
∫
π
/
2
0
sin
n
x
d
x
equals
n
−
1
n
n
−
3
n
−
2
n
−
5
n
−
4
.
.
.
1
2
×
π
2
Q.
If
A
n
=
∫
π
/
2
0
sin
(
2
n
−
1
)
x
sin
x
d
x
;
B
n
=
∫
π
/
2
0
(
sin
n
x
sin
x
)
2
d
x
;
for
n
∈
N
, then
Q.
If
I
=
∫
π
/
2
0
cos
n
x
sin
n
x
dx
=
k
∫
π
/
2
0
sin
n
x
dx
, then k equals