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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 3 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Indefinite Integrals
Quiz 3
∫
cos
x
log
(
cos
x
)
d
x
=
sin
x
log
(
cos
x
)
+
log
|
sec
x
+
tan
x
|
+
f
(
x
)
+
c
, then
f
(
x
)
=
Report Question
0%
−
sin
x
0%
cos
x
0%
tan
x
0%
cot
x
lf
∫
f
(
x
)
d
x
=
g
(
x
)
then
∫
x
3
f
(
x
2
)
d
x
=
Report Question
0%
1
2
{
x
2
g
(
x
2
)
−
∫
g
(
x
2
)
d
x
2
}
0%
1
2
{
x
2
g
(
x
2
)
−
∫
g
(
x
2
)
d
x
}
0%
1
2
{
x
2
g
(
x
2
)
+
∫
g
(
x
2
)
d
x
2
}
0%
1
2
{
x
2
g
(
x
2
)
+
∫
g
(
x
2
)
d
x
}
Explanation
∫
x
3
f
(
x
2
)
d
x
1
2
∫
x
2
f
(
x
2
)
d
x
2
x
2
=
k
1
2
∫
k
f
(
k
)
d
k
1
2
[
k
∫
f
(
k
)
d
k
−
∫
(
∫
f
(
k
)
d
k
)
d
k
]
=
1
2
[
k
g
(
k
)
−
∫
g
(
k
)
d
k
]
∫
f
(
x
)
d
x
=
∫
f
(
k
)
d
k
=
g
=
1
2
[
x
2
g
(
x
2
)
−
∫
g
(
x
2
)
d
x
2
]
=
1
2
[
x
2
g
(
x
2
)
−
∫
g
(
x
2
)
d
x
2
]
∫
{
f
(
x
)
g
″
(
x
)
−
f
″
(
x
)
g
(
x
)
}
d
x
=
Report Question
0%
f
(
x
)
g
′
(
x
)
−
g
(
x
)
f
′
(
x
)
+
c
0%
g
(
x
)
f
′
(
x
)
−
f
(
x
)
g
′
(
x
)
+
c
0%
g
(
x
)
f
′
(
x
)
+
f
(
x
)
g
′
(
x
)
+
c
0%
f
(
x
)
g
(
x
)
+
c
Explanation
∫
[
f
c
x
)
g
"
(
x
)
−
f
"
(
x
)
g
(
x
)
]
d
x
∫
f
(
x
)
g
″
(
x
)
+
f
′
(
x
)
g
′
(
x
)
d
x
−
∫
[
f
′
(
x
)
g
′
(
x
)
+
f
″
(
x
)
g
(
x
)
]
d
x
∫
d
[
f
c
x
)
g
′
(
x
)
]
−
∫
d
(
f
′
(
x
)
g
(
x
)
]
=
f
(
x
)
g
′
(
x
)
−
f
′
(
x
)
g
(
x
)
+
c
∫
[
f
(
x
)
g
″
(
x
)
−
f
″
(
x
)
g
(
x
)
]
d
x
=
f
(
x
)
g
′
(
x
)
−
f
′
(
x
)
g
(
x
)
+
c
∫
(
e
3
√
x
d
x
)
=
Report Question
0%
x
2
/
3
−
2
x
1
/
3
+
2
+
c
0%
(
x
2
/
3
−
2
x
1
/
3
+
2
)
exp
(
3
√
x
)
+
c
0%
3
(
x
2
/
3
−
2
x
1
/
3
+
2
)
exp
(
3
√
x
)
+
c
0%
2
(
x
2
/
3
−
2
x
1
/
3
+
2
)
exp
(
3
√
x
)
+
c
Explanation
I
=
∫
e
3
√
x
d
x
Put
3
√
x
=
t
1
3
x
−
2
/
3
d
x
=
d
t
d
x
=
3
t
−
2
d
t
So,
I
=
3
∫
t
2
e
t
d
t
=
3
[
t
2
e
t
−
2
∫
t
e
t
d
t
]
+
c
=
3
t
2
e
t
−
6
[
t
e
t
−
∫
e
t
d
t
]
+
c
=
3
t
2
e
t
−
6
t
e
t
+
6
e
t
+
c
=
3
(
x
2
/
3
−
2
x
1
/
3
+
2
)
e
3
√
x
+
c
∫
√
x
.
log
x
d
x
=
Report Question
0%
2
3
x
3
/
2
.
log
x
−
4
9
x
3
/
2
+
c
0%
2
3
x
3
/
2
.
log
x
+
x
3
/
2
+
c
0%
x
3
/
2
.
(
log
x
−
2
3
)
+
c
0%
2
5
x
3
/
2
(
log
x
+
1
)
+
c
Explanation
∫
√
x
l
o
g
x
d
x
=
l
o
g
x
∫
√
x
d
x
−
∫
1
/
x
(
∫
√
x
d
x
)
d
x
=
2
/
3
l
o
g
x
x
3
/
2
−
2
/
3
∫
x
3
/
2
x
d
x
2
/
3
l
o
g
x
(
x
3
/
2
)
−
2
/
3
∫
x
1
/
2
d
x
2
/
3
l
o
g
x
(
x
3
/
2
)
−
4
/
9
x
3
/
2
+
c
2
/
3
x
3
/
2
l
o
g
x
4
/
9
x
3
/
2
+
c
∫
√
x
l
o
g
x
d
x
=
2
/
3
x
3
/
2
l
o
g
x
−
4
/
9
x
3
/
2
+
c
lf
f
(
x
)
d
x
=
g
(
x
)
+
c
, then
∫
f
−
1
(
x
)
d
x
is equal to
Report Question
0%
x
f
−
1
(
x
)
+
c
0%
f
[
g
−
1
(
x
)
]
+
c
0%
x
f
−
1
(
x
)
−
g
[
f
−
1
(
x
)
]
+
c
0%
g
−
1
(
x
)
+
c
Explanation
∫
f
(
x
)
d
x
=
g
(
x
)
+
c
⇒
d
g
(
x
)
d
x
=
f
(
x
)
∫
f
−
1
(
x
)
d
x
let
x
=
f
(
y
)
⇒
∫
y
d
f
(
y
)
=
y
∫
d
f
(
y
)
−
∫
1
[
∫
d
f
(
y
)
]
d
y
=
y
f
(
y
)
−
∫
f
(
y
)
d
y
=
y
f
(
y
)
−
g
(
y
)
+
c
x
f
−
1
(
x
)
−
g
(
f
−
1
(
x
)
)
+
c
So,
∫
f
−
1
(
x
)
d
x
=
x
f
−
1
(
x
)
−
g
(
f
−
1
(
x
)
)
+
c
Evaluate
∫
e
sin
x
sin
2
x
d
x
Report Question
0%
2
e
sin
x
(
sin
x
+
1
)
+
c
0%
e
sin
x
(
sin
x
+
2
)
+
c
0%
e
sin
x
(
2
sin
x
−
2
)
+
c
0%
e
sin
x
(
3
sin
x
−
2
)
+
c
Explanation
∫
e
sin
x
⋅
2
sin
x
cos
x
d
x
=
2
∫
e
sin
x
⋅
sin
x
d
(
sin
x
)
Take
sin
x
=
t
=
2
∫
e
t
⋅
t
d
t
We know,
∫
u
.
v
d
x
=
u
∫
v
d
x
−
∫
(
d
u
d
x
∫
v
d
x
)
d
x
Now by integrating the functions by parts, by taking
t
as
u
and
e
t
as
v
, we get
=
2
(
t
e
t
−
e
t
)
=
2
e
t
(
t
−
1
)
Put the value of
t
=
2
e
sin
x
(
sin
x
−
1
)
=
e
sin
x
(
2
sin
x
−
2
)
⇒
∫
e
sin
x
sin
2
x
d
x
=
e
sin
x
(
2
sin
x
−
2
)
+
c
∫
x
tan
x
sec
2
x
d
x
=
Report Question
0%
1
2
[
x
tan
2
x
−
t
a
n
x
+
x
]
+
c
0%
1
2
[
x
tan
2
x
−
t
a
n
x
−
x
]
+
c
0%
1
2
[
x
tan
2
x
+
t
a
n
x
−
x
]
+
c
0%
x
tan
2
x
−
t
a
n
x
+
x
+
c
Explanation
L
=
∫
(
x
t
a
n
x
)
s
e
c
2
x
d
x
=
x
t
a
n
x
∫
s
e
c
2
x
d
x
−
∫
(
[
t
a
n
x
+
x
s
e
c
−
2
x
]
∫
s
e
c
2
x
d
x
)
d
x
=
x
t
a
n
x
t
a
n
x
−
∫
[
t
a
n
2
x
+
(
x
s
e
c
2
x
t
a
n
x
)
]
d
x
L
=
x
t
a
n
2
x
−
∫
t
a
n
2
x
d
x
−
∫
x
s
e
c
2
x
t
a
n
x
d
x
2
L
=
x
t
a
n
2
x
−
∫
(
s
e
c
2
x
−
1
)
d
x
2
L
=
x
t
a
n
2
x
−
t
a
n
x
+
x
+
c
L
=
1
2
[
x
t
a
n
2
x
−
t
a
n
x
+
x
]
+
c
∫
x
t
a
n
x
s
e
c
2
x
d
x
=
1
2
[
x
t
a
n
2
x
−
t
a
n
x
+
x
]
+
c
∫
2
x
+
s
i
n
2
x
1
+
c
o
s
2
x
d
x
=
Report Question
0%
x
s
i
n
2
x
+
c
0%
−
x
t
a
n
x
+
c
0%
x
s
e
c
x
+
c
0%
x
t
a
n
x
+
c
Explanation
∫
2
x
+
s
i
n
2
x
1
+
c
o
s
2
x
d
x
∫
2
x
+
s
i
n
2
x
2
c
o
s
2
x
d
x
∫
x
s
e
c
2
x
d
x
+
∫
t
a
n
x
d
x
x
∫
s
e
c
2
x
d
x
−
∫
(
∫
s
e
c
2
x
d
x
)
d
x
+
∫
t
a
n
x
d
x
x
∫
s
e
c
2
x
d
x
−
∫
t
a
n
x
d
x
+
∫
t
a
n
x
d
x
x
t
a
n
x
+
c
∫
2
x
+
s
i
n
2
x
1
+
c
o
s
2
x
=
x
t
a
n
x
+
c
∫
e
x
(
3
x
2
+
7
x
+
2
)
d
x
=
Report Question
0%
e
x
[
3
x
2
+
x
+
1
]
+
c
0%
e
x
[
3
x
2
+
7
x
+
1
]
+
c
0%
e
x
[
3
x
2
−
x
−
l
]
+
c
0%
e
x
[
3
x
2
+
x
]
+
c
Explanation
∫
e
x
(
3
x
2
+
7
x
+
2
)
d
x
3
∫
e
x
x
2
d
x
+
7
∫
e
x
x
d
x
+
2
∫
e
x
d
x
∫
e
x
x
2
d
x
=
x
2
e
x
−
2
∫
x
e
x
d
x
+
c
∫
x
e
x
=
x
e
x
−
e
x
+
c
∫
e
x
=
e
x
+
c
3
[
x
2
e
x
]
−
6
∫
x
e
x
d
x
+
7
∫
e
x
−
x
d
x
+
2
e
x
3
x
2
e
x
+
x
e
x
−
e
x
+
e
x
+
2
e
x
+
c
=
e
n
[
3
x
2
+
x
+
1
]
+
c
∫
e
x
(
3
x
2
+
7
x
+
2
)
d
x
=
e
x
[
3
x
2
+
x
+
1
]
+
c
Evaluate
∫
e
x
(
log
x
+
1
x
2
)
d
x
Report Question
0%
e
x
logx
+
c
0%
e
x
(
log
x
−
1
x
)
+
c
0%
e
x
(
log
x
+
1
x
)
+
c
0%
e
x
x
2
+
c
Explanation
∫
e
x
(
l
o
g
x
+
1
x
2
)
d
x
∫
e
x
l
o
g
x
d
x
+
∫
e
x
1
x
2
d
x
=
l
o
g
x
∫
e
x
d
x
−
∫
1
x
e
x
d
x
+
∫
e
x
x
2
d
x
=
l
o
g
x
e
x
−
∫
1
x
e
x
d
x
+
e
x
∫
1
x
2
d
x
−
∫
(
e
x
∫
1
x
2
d
x
)
d
x
=
l
o
g
x
e
x
−
∫
1
x
e
x
d
x
+
e
x
[
−
1
x
]
+
∫
e
x
1
x
d
x
=
e
x
[
l
o
g
x
−
1
x
]
+
c
So,
∫
e
x
(
l
o
g
x
+
1
x
2
)
d
x
=
e
x
[
l
o
g
x
−
1
x
]
+
c
∫
x
3
cos
x
2
d
x
=
Report Question
0%
x
2
sin
x
2
−
cos
x
2
+
c
0%
1
2
[
x
2
sin
x
2
+
cos
x
2
]
+
c
0%
1
3
[
x
2
sin
x
2
+
cos
x
2
]
+
c
0%
x
2
sin
x
2
+
cos
x
2
+
c
Explanation
∫
x
3
cos
x
2
d
x
Put
x
2
=
t
x
d
x
=
d
t
2
=
1
2
∫
t
cos
t
d
t
=
1
2
[
t
sin
t
−
∫
sin
t
d
t
]
=
1
2
t
sin
t
+
1
2
cos
t
+
c
=
1
2
[
x
2
sin
x
2
+
cos
x
2
]
+
c
∫
log
(
x
2
+
a
2
)
x
2
d
x
=
Report Question
0%
−
log
(
x
2
+
a
2
)
x
+
2
a
t
a
n
−
1
(
x
a
)
+
c
0%
log
(
x
2
+
a
2
)
x
+
2
a
t
a
n
−
1
(
x
a
)
+
c
0%
log
(
x
2
+
a
2
)
x
−
2
a
t
a
n
−
1
(
x
a
)
+
c
0%
−
log
(
x
2
+
a
2
)
x
−
2
a
t
a
n
−
l
(
x
a
)
+
c
Explanation
∫
l
o
g
(
x
2
+
a
2
)
x
2
d
x
l
o
g
(
x
2
+
a
2
)
∫
1
/
x
2
d
x
−
∫
(
2
x
x
2
+
a
2
−
[
∫
1
x
2
d
x
]
)
d
x
=
l
o
g
(
x
2
+
a
2
)
(
−
1
/
x
)
+
2
∫
1
(
x
2
+
a
2
)
d
x
+
c
=
l
o
g
(
x
2
+
a
2
)
(
−
1
/
x
)
+
2
a
t
a
n
−
1
(
x
/
a
)
+
c
=
−
l
o
g
(
x
2
+
a
2
)
x
+
2
a
t
a
n
−
1
(
x
/
a
)
+
c
∫
l
o
g
(
x
2
+
a
2
)
x
2
d
x
=
−
l
o
g
(
x
2
+
a
2
)
x
+
2
a
t
a
n
−
1
(
x
/
a
)
+
c
∫
x
sin
−
1
x
√
1
−
x
2
d
x
is equal to
Report Question
0%
x
−
√
1
−
x
2
s
i
n
−
1
x
+
c
0%
x
+
√
1
−
x
2
s
i
n
−
1
x
+
c
0%
x
+
s
i
n
−
1
x
+
c
0%
x
−
s
i
n
−
1
x
+
c
Explanation
Let
I
=
∫
x
sin
−
1
x
√
1
−
x
2
d
x
Put
sin
−
1
x
=
t
⟹
x
=
sin
t
1
√
1
−
x
2
d
x
=
d
t
I
=
∫
t
sin
t
d
t
=
−
t
cos
t
+
∫
cos
t
d
t
....integration by parts
=
−
t
cos
t
+
sin
t
+
C
=
−
√
1
−
x
2
sin
−
1
x
+
x
+
C
=
x
−
√
1
−
x
2
sin
−
1
x
+
C
Hence, option 'A' is correct.
∫
e
s
i
n
−
1
x
[
1
+
x
√
1
−
x
2
]
dx
=
Report Question
0%
x
e
s
i
n
−
1
x
+
c
0%
e
s
i
n
−
1
x
+
c
0%
1
√
1
−
x
2
e
sin
−
1
x
+
c
0%
x
2
e
s
i
n
1
X
+
c
Explanation
I
=
∫
e
s
i
n
−
1
x
[
1
+
x
√
1
−
x
2
]
d
x
Put
sin
−
1
x
=
t
⇒
x
=
sin
t
⇒
d
x
=
cos
t
d
t
So,
I
=
∫
e
t
(
sin
t
+
cos
t
)
d
t
⇒
I
=
(
sin
t
+
cos
t
)
e
t
−
∫
(
cos
t
−
sin
t
)
e
t
d
t
⇒
I
=
(
sin
t
+
cos
t
)
e
t
−
(
cos
t
−
sin
t
)
e
t
−
∫
e
t
(
sin
t
+
cos
t
)
d
t
⇒
2
I
=
2
sin
t
e
t
+
C
⇒
I
=
x
e
sin
−
1
x
+
C
∫
e
√
x
d
x
=
Report Question
0%
2
e
√
x
(
√
x
+
1
)
+
c
0%
2
e
√
x
(
√
x
−
1
)
+
c
0%
e
x
(
x
−
1
)
+
c
0%
e
x
(
x
+
1
)
+
c
Explanation
I
=
∫
e
√
x
d
x
Put
√
x
=
t
1
2
x
−
1
/
2
d
x
=
d
t
⇒
d
x
=
2
t
d
t
So,
I
=
2
∫
t
e
t
d
t
=
2
[
t
e
t
−
∫
e
t
d
t
]
=
2
t
e
t
−
2
e
t
+
C
=
2
e
√
x
(
√
x
−
1
)
+
c
∫
t
a
n
−
1
√
1
−
x
1
+
x
d
x
=
Report Question
0%
1
2
[
x
c
o
s
−
1
x
−
√
1
−
x
2
]
+
c
0%
[
x
c
o
s
−
1
x
−
√
1
−
x
2
]
+
c
0%
1
2
[
x
c
o
s
−
1
x
+
√
1
−
x
2
]
+
c
0%
x
c
o
s
−
1
x
+
√
1
−
x
2
+
c
Explanation
I
=
∫
t
a
n
−
1
√
1
−
x
1
+
x
d
x
=
Put
x
=
cos
θ
⇒
d
x
=
−
sin
θ
d
θ
I
=
∫
t
a
n
−
1
(
tan
θ
2
)
(
−
sin
θ
)
d
θ
⇒
I
=
−
∫
θ
2
sin
θ
d
θ
⇒
I
=
−
1
2
∫
θ
sin
θ
d
θ
⇒
I
=
−
1
2
[
θ
(
−
cos
θ
)
+
sin
θ
]
+
c
⇒
I
=
1
2
[
θ
(
cos
θ
)
−
sin
θ
]
+
c
I
=
1
2
[
x
c
o
s
−
1
x
−
√
1
−
x
2
]
+
c
(A) :
∫
(
2
x
tan x
sec
2
x
+
tan
2
x
)
d
x
=
x
tan
2
x
+
c
(B) :
∫
[
x
f
′
(
x
)
+
f
(
x
)
]
d
x
=
x
f
(
x
)
+
c
Report Question
0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true but R is not correct explanation of A
0%
A is true R is false
0%
A is false but R is true.
Explanation
B is true
A is in the form of B
So, A and B are true.
and B is correct explanation.
∫
x
−
sin
x
1
−
cos
x
d
x
=
Report Question
0%
l
o
g
|
1
−
C
o
s
x
|
+
c
0%
l
o
g
|
x
−
s
i
n
x
|
+
c
0%
x
tan
x
2
+
c
0%
−
x
cot
x
2
+
c
Explanation
∫
(
x
−
s
i
n
x
1
−
c
o
s
x
)
d
x
=
∫
x
−
s
i
n
x
2
s
i
n
2
x
/
2
d
x
=
∫
(
x
2
)
c
o
s
e
c
2
x
/
2
d
x
−
∫
c
o
t
x
/
2
d
x
=
x
/
2
∫
c
o
s
e
c
2
x
/
2
d
(
x
/
2
−
∫
(
∫
c
o
s
e
c
2
x
/
2
d
x
/
2
)
)
d
x
−
∫
c
o
t
x
/
2
d
x
=
−
x
c
o
t
x
/
2
+
c
+
∫
c
o
t
x
/
2
d
x
−
∫
c
o
t
x
/
2
d
x
=
−
x
c
o
t
x
/
2
+
c
∫
(
x
−
s
i
n
x
1
−
c
o
s
x
)
d
x
=
−
x
c
o
t
x
/
2
+
c
∫
(
log
x
)
2
d
x
=
Report Question
0%
x
[
(
log
x
)
2
−
2
log
x
+
2
]
+
c
0%
x
[
(
log
x
)
2
+
2
log
x
+
2
]
+
c
0%
[
(
log
x
)
2
−
2
log
x
+
2
]
+
c
0%
[
(
log
x
)
2
+
2
log
x
+
2
]
+
c
Explanation
∫
(
log
x
)
2
1
d
x
∫
(
log
x
)
2
−
1
d
x
(
log
x
)
2
∫
1
d
x
−
∫
2
(
log
x
)
x
x
d
x
=
(
log
x
)
2
[
x
]
−
2
∫
log
x
d
x
∫
log
x
=
x
log
x
−
x
=
x
(
log
x
)
2
−
2
[
x
log
x
−
x
]
+
c
=
x
(
log
x
)
2
−
2
x
(
log
x
)
+
2
x
+
c
=
x
[
(
log
x
)
2
−
2
(
log
x
)
+
2
]
+
c
⇒
∫
(
log
x
)
2
d
x
=
x
[
(
log
x
)
2
−
2
(
log
x
)
+
2
]
+
c
∫
x
n
.
log
x
d
x
=
Report Question
0%
x
n
+
1
(
n
+
1
)
2
(
log
x
−
1
(
n
+
1
)
2
)
+
c
0%
(
log
x
−
1
(
n
+
1
)
2
)
+
c
0%
x
n
+
1
(
n
+
1
)
(
log
x
−
1
(
n
+
1
)
)
+
c
0%
x
n
+
1
(
n
+
1
)
(
log
x
−
1
(
n
+
1
)
2
)
+
c
Explanation
∫
x
n
l
o
g
x
d
x
=
l
o
g
x
∫
x
n
d
x
−
∫
[
1
x
∫
x
n
.
d
x
]
d
x
=
l
o
g
x
x
n
+
1
n
+
1
−
1
n
+
1
∫
x
n
.
d
x
=
l
o
g
x
x
n
+
1
n
+
1
−
1
n
+
1
[
x
n
+
1
(
n
+
1
)
]
+
c
=
x
n
+
1
(
n
+
1
)
[
l
o
g
x
−
1
(
n
+
1
)
]
+
c
∫
x
n
l
o
g
x
d
x
=
x
n
+
1
n
+
1
[
l
o
g
x
−
1
(
n
+
1
)
]
+
c
∫
log
(
1
−
x
)
x
2
d
x
equal to
Report Question
0%
[
(
1
+
1
x
)
log
(
1
−
x
)
−
log
x
]
+
c
0%
(
1
−
1
x
)
log
(
1
−
x
)
−
log
x
+
c
0%
(
1
x
−
1
)
log
(
1
−
x
)
+
log
x
+
c
0%
(
1
x
−
1
)
log
(
1
−
x
)
−
log
x
+
c
Explanation
∫
l
o
g
(
1
−
x
)
x
2
d
x
l
o
g
(
1
−
x
)
∫
1
/
x
2
d
x
−
∫
[
1
(
1
−
x
)
∫
1
x
2
d
x
]
d
x
−
l
o
g
(
1
−
x
)
x
+
c
−
∫
1
(
1
−
x
)
−
1
x
d
x
=
−
l
o
g
(
1
−
x
)
x
−
∫
[
1
x
+
1
1
−
x
]
d
x
+
c
=
−
l
o
g
(
1
−
x
)
x
−
l
o
g
x
+
l
o
g
(
1
−
x
)
+
c
=
l
o
g
(
1
−
x
)
[
1
−
1
x
]
−
l
o
g
x
+
c
∫
l
o
g
(
1
−
x
)
x
2
d
x
−
l
o
g
(
1
−
x
)
[
1
−
1
/
x
]
−
l
o
g
x
+
c
∫
sin
√
x
d
x
=
Report Question
0%
2
(
S
i
n
√
x
−
√
x
.
cos
√
x
)
+
c
0%
2
(
S
i
n
√
x
+
√
x
.
cos
√
x
)
+
c
0%
2
(
S
i
n
√
x
−
1
2
√
x
.
cos
√
x
)
+
c
0%
2
(
S
i
n
√
x
+
1
2
√
x
.
cos
√
x
)
+
c
Explanation
∫
32
x
3
(
log
x
)
2
d
x
is equal to
Report Question
0%
8
x
4
(
log
x
)
2
+
c
0%
x
4
{
8
(
log
x
)
2
−
4
log
x
+
1
}
+
c
0%
x
4
{
8
(
log
x
)
2
−
4
log
x
}
+
c
0%
x
3
{
(
log
x
)
2
+
2
log
x
}
+
c
Explanation
∫
32
x
3
(
l
o
g
x
)
2
=
32
∫
x
3
(
l
o
g
x
)
2
d
x
=
32
x
3
∫
(
l
o
g
x
)
[
(
l
o
g
x
)
2
]
x
3
d
x
−
∫
(
2
l
o
g
x
x
x
4
4
)
d
x
=
(
l
o
g
x
)
2
−
x
4
4
−
∫
1
/
2
l
o
g
x
.
x
3
−
d
x
=
(
l
o
g
x
)
2
−
x
4
4
−
1
/
2
∫
l
o
g
x
−
x
3
−
d
x
∫
l
o
g
x
.
x
3
d
x
=
l
o
g
x
.
x
4
4
−
∫
1
/
x
.
x
4
3
4
.
d
x
=
l
o
g
x
.
x
4
4
−
1
/
16
x
4
+
c
32
[
(
l
o
g
x
)
2
−
x
4
4
−
l
o
g
x
.
x
4
8
+
1
32
x
4
]
+
c
=
x
4
[
8
(
l
o
g
x
)
2
−
4
l
o
g
x
+
1
]
+
c
∫
x
3
(
log
x
)
2
d
x
=
Report Question
0%
x
4
4
[
(
log
x
)
2
−
log
x
2
+
1
8
]
+
c
0%
x
4
4
[
(
log
x
)
2
+
log
x
2
+
1
8
]
+
c
0%
(
log
x
)
2
+
2
l
o
g
x
+
8
+
c
0%
x
4
4
[
(
log
x
)
2
−
log
x
2
−
1
8
]
+
c
Explanation
∫
x
3
(
l
o
g
x
)
2
d
x
=
(
l
o
g
x
)
2
−
∫
x
3
d
x
−
∫
(
2
l
o
g
x
x
.
x
4
4
)
d
x
=
(
l
o
g
x
)
2
−
x
4
4
−
∫
1
2
(
l
o
g
x
)
(
x
3
)
d
x
∫
(
l
o
g
x
)
(
x
3
)
d
x
=
l
o
g
x
.
x
4
4
−
x
4
16
+
c
x
4
4
[
(
l
o
g
x
)
2
−
l
o
g
x
2
+
1
8
]
+
c
∫
s
i
n
x
2
+
c
o
s
2
x
2
1
+
c
o
s
x
d
x
=
Report Question
0%
sec
x
2
+
x
+
c
0%
sec
x
2
+
x
2
+
c
0%
sec
x
2
−
x
2
+
C
0%
sec
x
2
−
X
+
C
Explanation
∫
s
i
n
x
2
+
c
o
s
2
x
2
1
+
c
o
s
x
d
x
∫
s
i
n
x
2
+
c
o
s
2
x
2
2
c
o
s
2
x
2
d
x
=
x
2
[
∫
t
a
n
x
2
s
e
c
x
2
d
x
2
+
∫
1.
d
x
]
s
e
c
x
2
+
x
+
c
Solve
∫
e
log
x
.
cos
x
d
x
Report Question
0%
x
sin
x
−
c
o
s
x
+
c
0%
x
2
sin
x
+
cos
x
+
c
0%
x
sin
x
+
c
o
s
x
+
c
0%
x
sin
x
+
c
o
s
2
x
+
c
Explanation
We know,
e
l
o
g
x
=
x
∫
e
l
o
g
x
c
o
s
x
d
x
=
∫
x
c
o
s
x
d
x
We know,
∫
u
.
v
d
x
=
u
∫
v
d
x
−
∫
(
d
u
d
x
∫
v
d
x
)
d
x
Now by integrating the functions by parts, taking
x
as
u
and
c
o
s
x
as
v
, we get
∫
x
c
o
s
x
d
x
=
x
∫
c
o
s
x
d
x
−
∫
1
(
∫
c
o
s
x
)
d
x
=
x
s
i
n
x
+
c
o
s
x
+
c
∫
e
l
o
g
x
c
o
s
x
d
x
=
x
⋅
s
i
n
x
+
c
o
s
x
+
c
⋅
∫
cos
x
.
log
(
cos
x
)
d
x
=
Report Question
0%
s
i
n
x
l
o
g
(
c
o
s
x
)
−
l
o
g
(
c
o
s
x
)
+
c
0%
s
i
n
x
l
o
g
(
c
o
s
x
)
+
s
e
c
x
+
c
0%
s
i
n
x
l
o
g
(
c
o
s
x
)
−
s
i
n
x
+
l
o
g
|
s
e
c
x
+
t
a
n
x
|
+
c
0%
s
i
n
x
l
o
g
(
c
o
s
x
)
−
s
e
c
x
+
c
Explanation
∫
c
o
s
x
l
o
g
(
c
o
s
x
)
d
x
.
=
l
o
g
(
c
o
s
x
)
∫
c
o
s
x
d
x
−
∫
[
l
o
g
(
c
o
s
x
)
d
x
−
∫
(
c
o
s
x
)
d
x
]
=
l
o
g
(
c
o
s
x
)
s
i
n
x
−
∫
(
−
s
i
n
x
)
.
(
t
s
i
n
x
)
c
o
s
x
d
x
=
l
o
g
(
c
o
s
x
)
s
i
n
x
+
∫
s
i
n
2
x
c
o
s
x
d
x
=
l
o
g
(
c
o
s
x
)
s
i
n
x
+
∫
s
e
c
x
d
x
−
∫
c
o
s
x
d
x
=
s
i
n
x
l
o
g
(
c
o
s
x
)
−
s
i
n
x
+
l
o
g
|
s
e
c
x
+
t
a
n
x
|
+
c
∫
x
3
s
i
n
x
2
d
x
=
Report Question
0%
1
2
sin
x
2
−
1
2
x
2
cos
x
2
+
c
0%
1
2
sin
x
2
+
1
2
x
2
cos
x
2
+
c
0%
sin
x
2
−
x
2
cos
x
2
+
c
0%
sin
x
2
−
1
2
x
2
cos
x
2
+
c
Explanation
∫
x
3
s
i
n
x
2
d
x
=
1
/
2
∫
x
2
s
i
n
x
2
d
x
2
,
l
e
t
x
2
=
t
=
1
/
2
∫
t
s
i
n
t
d
t
∫
t
s
i
n
t
d
t
=
−
t
c
o
s
t
+
s
i
n
t
=
1
/
2
∫
t
s
i
n
d
t
=
1
/
2
[
s
i
n
t
−
t
c
o
s
t
]
+
c
=
1
/
2
[
s
i
n
x
2
−
x
2
c
o
s
x
2
]
+
c
=
1
/
2
s
i
n
x
2
−
x
2
/
2
c
o
s
x
2
+
c
∫
x
3
s
i
n
x
2
d
x
=
1
/
2
s
i
n
x
2
−
x
2
2
c
o
s
x
2
+
c
Evaluate
∫
x
.
sec
2
x
d
x
Report Question
0%
x
tan
x
+
log
|
sec
x
|
+
c
0%
x
tan
x
−
log
|
sec
x
|
+
c
0%
x
2
tan
x
−
log
|
sec
x
|
+
c
0%
x
2
tan
x
+
log
|
sec
x
|
+
c
Explanation
We know,
∫
u
.
v
d
x
=
u
∫
v
d
x
−
∫
(
d
u
d
x
∫
v
d
x
)
d
x
Now by integrating the functions by parts, taking
x
as
u
and
s
e
c
2
x
as
v
.
∫
x
.
sec
2
x
d
x
=
x
∫
sec
2
x
d
x
−
∫
1
(
∫
sec
2
x
d
x
)
d
x
=
x
tan
x
−
∫
tan
x
d
x
+
c
=
x
tan
x
−
log
|
sec
x
|
+
c
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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