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Standard XII
Mathematics
Integration of Trigonometric Functions
Question
∫
sin
−
1
(
2
x
1
+
x
2
)
d
x
=
f
(
x
)
−
log
(
1
+
x
2
)
+
c
then
f
(
x
)
=
2
x
tan
−
1
x
x
tan
−
1
x
−
2
x
tan
−
1
x
−
x
tan
−
1
x
A
x
tan
−
1
x
B
−
x
tan
−
1
x
C
2
x
tan
−
1
x
D
−
2
x
tan
−
1
x
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Solution
Verified by Toppr
∫
s
i
n
−
1
(
2
x
1
+
x
2
)
d
x
=
f
(
x
)
−
l
o
g
(
1
+
x
2
)
+
c
∫
s
i
n
−
1
(
2
x
1
+
x
2
)
d
x
;
l
e
t
x
=
t
a
n
θ
∫
s
i
n
−
1
(
2
t
a
n
θ
1
+
t
a
n
2
θ
)
s
e
c
2
θ
d
θ
=
2
∫
θ
s
e
c
2
θ
d
o
=
2
[
θ
∫
s
e
c
2
θ
d
θ
−
∫
1
(
∫
s
e
c
2
θ
d
θ
)
d
θ
]
=
2
[
θ
t
a
n
θ
−
∫
t
a
n
θ
d
θ
+
c
]
2
θ
t
a
n
θ
+
2
l
o
g
|
c
o
s
θ
|
+
c
converting in to 'x'
2
x
t
a
n
−
1
x
+
2
(
l
o
g
1
√
1
+
x
2
)
+
c
f
(
x
)
=
2
x
t
a
n
−
1
x
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14
Similar Questions
Q1
f
(
x
)
=
2
x
tan
−
1
x
,
g
(
x
)
=
log
(
1
+
x
2
)
,
h
(
x
)
=
sin
x
,
u
(
x
)
=
x
−
x
3
6
+
x
4
120
then
View Solution
Q2
The integral of
tan
−
1
(
x
)
is
-
View Solution
Q3
If
sin
−
1
2
x
1
+
x
2
;
cos
−
1
1
−
x
2
1
+
x
2
;
tan
−
1
2
x
1
−
x
2
.
each is equal to
2
tan
−
1
x
.
,then show that
∫
2
tan
−
1
x
=
2
[
x
tan
−
1
x
−
1
2
log
(
1
+
x
2
)
]
View Solution
Q4
Solve
∫
x
tan
−
1
x
1
+
x
2
d
x
View Solution
Q5
∫
∞
0
x
tan
−
1
x
(
1
+
x
2
)
d
x
View Solution