Byju's Answer
Standard XIII
Mathematics
Property 1
∫0π/2cos x/1+...
Question
∫
π
2
0
c
o
s
x
(
1
+
s
i
n
x
)
(
2
+
s
i
n
x
)
d
x
=
A
l
o
g
4
3
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B
l
o
g
1
3
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C
l
o
g
3
4
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D
N
o
n
e
o
f
t
h
e
s
e
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Solution
The correct option is
A
l
o
g
4
3
Put sinx=t
⇒
c
o
s
x
d
x
=
d
t
,
so that reduced integral is
∫
1
0
(
1
1
+
t
−
1
2
+
t
)
d
t
=
[
l
o
g
(
1
+
t
)
−
l
o
g
(
2
+
t
)
]
1
0
=
l
o
g
2
3
−
l
o
g
1
2
=
l
o
g
4
3
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0
Similar questions
Q.
∫
π
2
0
c
o
s
x
(
1
+
s
i
n
x
)
(
2
+
s
i
n
x
)
d
x
=
Q.
∫
π
2
0
c
o
s
x
(
1
+
s
i
n
x
)
(
2
+
s
i
n
x
)
d
x
=
[UPSEAT 1999]
Q.
∫
π
2
0
c
o
x
x
−
s
i
n
x
1
+
s
i
n
x
c
o
s
x
d
x
=
Q.
∫
π
0
x
s
i
n
x
1
+
c
o
s
2
x
d
x
=
Q.
If the reduction formula for
I
n
=
∫
sin
nx
cos
x
dx
is given by
I
n
+
I
n
−
2
=
−
2
cos
{
(
n
−
1
)
x
}
n
−
1
, then
∫
sin
3
x
cos
x
dx
is.
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