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Byju's Answer
Standard X
Physics
Magnetic Field Due to Straight Current Carrying Conductor
∫sin x+cos x/...
Question
∫
sin
x
+
cos
x
√
1
+
sin
2
x
d
x
=
A
x
+
c
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B
−
x
+
c
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C
sin
x
−
c
o
s
x
+
c
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D
|
x
|
+
c
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Solution
The correct option is
C
|
x
|
+
c
∫
sin
x
+
cos
x
√
1
+
sin
2
x
d
x
∫
sin
x
+
cos
x
√
1
+
2
sin
x
.
cos
x
d
x
=
∫
sin
x
+
cos
x
√
sin
2
x
+
cos
2
+
2
sin
x
.
cos
x
d
x
=
∫
sin
x
+
cos
x
√
(
sin
x
+
cos
x
)
2
d
x
=
∫
sin
x
+
cos
x
|
sin
x
+
cos
x
|
d
x
=
∫
|
1
|
d
x
=
|
x
|
+
c
Suggest Corrections
0
Similar questions
Q.
Solve
∫
sin
x
+
cos
x
√
1
+
sin
2
x
d
x
, Given that x takes values for
sin
x
+
cos
x
≥
0
.
Q.
Let
I
=
∫
sin
2
x
+
sin
x
1
+
sin
x
+
cos
x
d
x
,
J
=
∫
cos
2
x
+
cos
x
1
+
sin
x
+
cos
x
d
x
and
c
is the constant of integration.
Function
Integral
(a) I
(p)
1
2
(
x
−
sin
x
−
cos
x
)
+
c
(b) J
(q)
1
2
(
x
+
sin
x
+
cos
x
)
+
c
(c) I + J
(r)
x
+
c
(d) I - J
(s)
c
−
cos
x
−
sin
x
(t)
c
+
cos
x
+
sin
x
(u)
−
1
2
(
x
+
sin
x
+
cos
x
+
c
)
Then the value of
d
(
I
+
J
)
d
x
at
x
=
√
2
is
Q.
Show that
∫
sin
x
+
cos
x
√
(
1
+
sin
2
x
)
d
x
=
x
.
Q.
∫
2
s
i
n
x
(
3
+
s
i
n
2
x
)
d
x
is equal to
Q.
I:
∫
cos
(
x
−
a
)
sin
x
d
x
=
cos
a
log
|
sin
x
|
+
x
sin
a
+
c
II:
∫
sin
(
x
−
a
)
cos
(
x
−
b
)
d
x
=
cos
(
b
−
a
)
log
|
sec
(
x
−
b
)
|
+
x
sin
(
b
−
a
)
+
c
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