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Byju's Answer
Standard XII
Mathematics
Integration Using Substitution
∫0π /4√tanθ d...
Question
∫
π
/
4
0
√
tan
θ
d
θ
=
1
√
(
2
)
l
o
g
(
√
2
−
1
)
+
π
2
k
√
(
2
)
.Find the value of
k
.
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Solution
Let
I
=
∫
√
tan
θ
d
θ
=
1
2
(
I
1
−
I
2
)
Where
I
1
=
∫
(
tan
θ
+
cot
θ
)
d
θ
=
∫
sin
θ
+
cos
θ
√
cos
θ
sin
θ
d
θ
Put
sin
θ
+
cos
θ
=
u
I
1
=
√
2
∫
d
t
√
1
−
t
2
=
√
2
sin
−
1
t
=
√
2
sin
−
1
(
sin
θ
−
cos
θ
)
And
I
2
=
∫
(
√
cot
θ
−
√
tan
θ
)
d
θ
=
∫
cos
θ
−
sin
θ
√
cos
θ
sin
θ
d
θ
Put
sin
θ
+
cos
θ
=
t
⇒
2
sin
θ
cos
θ
=
t
2
−
1
I
2
=
√
2
∫
d
t
√
t
2
−
1
=
√
2
log
(
1
+
√
t
2
−
1
)
=
√
2
log
(
sin
θ
+
cos
θ
+
√
sin
2
θ
)
I
=
√
2
sin
−
1
(
sin
θ
−
cos
θ
)
+
√
2
log
(
sin
θ
+
cos
θ
+
√
sin
2
θ
)
therefore
∫
π
/
4
0
I
d
θ
=
√
2
log
(
√
2
+
1
)
+
π
2
√
2
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0
Similar questions
Q.
The value of
∫
π
/
4
0
√
tan
θ
d
θ
is?
Q.
Show that
π
2
∫
0
x
sin
x
+
cos
x
d
x
=
π
2
√
2
log
(
√
2
+
1
)
Q.
∫
π
/
2
0
cos
x
d
x
1
−
sin
2
x
+
sin
4
x
=
π
k
+
1
4
√
(
3
)
l
o
g
(
(
2
+
√
3
)
(
2
−
√
3
)
)
. Find the value of
k
.
Q.
∫
1
sec
x
+
c
o
sec
x
d
x
=
1
2
[
−
cos
x
+
sin
x
]
−
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2
√
2
log
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+
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. Find the value of
k
.
Q.
∫
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+
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d
x
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