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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 2 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Indefinite Integrals
Quiz 2
∫
sin
−
1
(
cos
x
)
d
x
is equal to
Report Question
0%
π
x
2
+
c
0%
π
x
2
2
+
c
0%
π
x
−
x
2
2
+
c
0%
π
x
+
x
2
2
+
c
Explanation
sin
−
1
(
cos
x
)
=
π
2
−
cos
−
1
(
cos
x
)
=
π
2
−
x
∫
sin
−
1
(
cos
x
)
d
x
=
∫
(
π
2
−
x
)
d
x
=
π
x
−
x
2
2
+
C
What is
∫
d
x
x
(
1
+
l
n
x
)
n
equal to
(
n
≠
1
)
?
Report Question
0%
1
(
n
−
1
)
(
1
+
l
n
x
)
n
−
1
+
c
0%
1
−
n
(
1
+
l
n
x
)
1
−
n
+
c
0%
n
+
1
(
1
+
l
n
x
)
n
+
1
+
c
0%
−
1
(
n
−
1
)
(
1
+
l
n
x
)
n
−
1
+
c
Explanation
Given,
∫
1
x
(
1
+
ln
(
x
)
)
n
d
x
apply
u
=
1
+
ln
(
x
)
=
∫
1
u
n
d
u
=
∫
u
−
n
d
u
=
u
−
n
+
1
−
n
+
1
=
(
1
+
ln
(
x
)
)
−
n
+
1
−
n
+
1
=
(
1
+
ln
(
x
)
)
−
n
+
1
−
n
+
1
+
C
−
1
(
n
−
1
)
(
1
+
l
n
x
)
n
−
1
+
c
Evaluate
∫
x
s
e
c
x
.
t
a
n
x
d
x
=
Report Question
0%
x
sec
x
+
log
|
tan
(
π
/
2
+
x
/
2
)
|
+
c
0%
x
sec
x
−
log
|
tan
(
π
/
4
+
x
/
2
)
|
+
c
0%
x
sec
x
−
log
|
tan
(
π
/
4
+
x
)
|
+
c
0%
x
sec
x
+
log
|
tan
x
/
2
|
+
c
Explanation
∫
x
sec
x
tan
x
d
x
=
∫
x
(
sec
x
(
tan
x
)
)
d
x
=
x
∫
(
sec
x
tan
x
)
d
x
−
∫
1
∫
(
sec
x
tan
x
)
d
x
....... [Using integration parts]
=
x
sec
x
−
∫
sec
x
d
x
=
x
sec
x
−
log
|
sec
x
+
tan
x
|
+
c
=
x
sec
x
−
log
|
tan
(
π
/
4
+
x
/
2
)
|
+
c
.....
[
∵
sec
x
+
tan
x
=
tan
(
π
/
4
+
x
/
2
)
]
∴
∫
x
sec
x
tan
x
d
x
=
x
sec
x
−
log
|
tan
(
π
/
4
+
x
/
2
)
|
+
c
∫
{
(
log
x
−
1
)
1
+
(
log
x
)
2
}
2
d
x
is equal to
Report Question
0%
x
(
log
x
)
2
+
1
+
c
0%
x
e
x
1
+
x
2
+
c
0%
x
x
2
+
1
+
c
0%
log
x
(
log
x
)
2
+
1
+
c
Explanation
∫
[
l
o
g
x
−
1
1
+
(
l
o
g
x
)
2
]
2
d
x
∫
(
l
o
g
x
−
1
)
2
[
1
+
(
l
o
g
x
)
2
]
2
=
∫
(
l
o
g
x
)
2
+
1
−
2
l
o
g
x
)
[
1
+
(
l
o
g
x
)
2
]
2
=
∫
(
1
[
(
1
+
(
l
o
g
x
)
2
]
−
2
l
o
g
x
[
1
+
l
o
g
2
x
]
2
)
d
x
=
∫
1
(
1
+
l
o
g
2
x
)
d
x
−
∫
2
l
o
g
x
(
1
+
l
o
g
2
x
)
2
d
x
=
1
(
1
+
l
o
g
2
x
)
∫
1
d
x
+
∫
[
2
l
o
g
x
(
1
+
l
o
g
2
x
)
2
×
1
x
∫
1
⋅
d
x
]
d
x
−
∫
2
l
o
g
x
(
1
+
l
o
g
2
x
)
2
d
x
=
x
1
+
l
o
g
2
x
+
∫
2
l
o
g
x
(
1
+
l
o
g
2
x
)
2
d
x
−
∫
2
l
o
g
x
(
1
+
l
o
g
2
x
)
2
d
x
+
c
=
x
1
+
l
o
g
2
x
+
c
∫
sec
2
x
.
cosec
2
x
d
x
=
Report Question
0%
tan
x
−
cot
x
+
c
0%
tan
x
+
cot
x
+
c
0%
−
tan
x
+
cot
x
+
c
0%
sec
x
tan
x
+
c
Explanation
∫
sec
2
x
.
cosec
2
x
d
x
=
∫
1
cos
2
x
.
sin
2
x
d
x
=
∫
cos
2
x
+
sin
2
x
cos
2
x
.
sin
2
x
d
x
=
∫
cos
2
x
cos
2
x
sin
2
x
+
sin
2
x
cos
2
x
.
sin
2
x
d
x
=
∫
1
sin
2
x
+
1
cos
2
x
d
x
=
∫
(
cosec
2
x
+
sec
2
x
)
d
x
=
−
cot
x
+
tan
x
+
c
=
tan
x
−
cot
x
+
c
∫
√
x
x
−
1
d
x
,
x
∈
(
0
,
π
/
2
)
equals
Report Question
0%
√
x
(
x
−
1
)
+
log
(
√
x
+
√
x
−
1
)
+
c
0%
√
x
(
x
−
1
)
−
log
(
√
x
+
√
x
−
1
)
+
c
0%
√
x
(
x
−
1
)
+
log
(
√
x
−
√
x
−
1
)
+
c
0%
√
x
(
x
+
1
)
−
log
(
√
x
−
√
x
+
1
)
+
c
Explanation
x
=
s
e
c
2
θ
d
x
=
2
s
e
c
2
θ
⋅
t
a
n
θ
⋅
d
θ
∫
√
s
e
c
2
θ
t
a
n
2
θ
⋅
2
s
e
c
2
θ
t
a
n
θ
d
θ
∫
1
s
i
n
θ
⋅
2
s
e
c
2
θ
⋅
s
i
n
θ
c
o
s
θ
⋅
d
θ
=
2
∫
s
e
c
3
θ
⋅
d
θ
⋅
=
2
∫
(
s
e
c
θ
)
(
s
e
c
2
θ
)
d
θ
⋅
=
2
[
s
e
c
θ
t
a
n
θ
−
∫
(
s
e
c
θ
t
a
n
θ
⋅
t
a
n
θ
)
d
θ
⋅
]
2
∫
s
e
c
3
θ
d
θ
=
2
[
s
e
c
θ
t
a
n
θ
]
−
2
∫
s
e
c
θ
(
s
e
c
2
θ
−
1
)
d
θ
2
∫
s
e
c
3
θ
d
θ
=
s
e
c
θ
t
a
n
θ
+
∫
s
e
c
θ
d
θ
2
∫
s
e
c
3
θ
d
θ
=
s
e
c
θ
t
a
n
θ
→
l
o
g
|
s
e
c
θ
+
t
a
n
θ
|
and
=
√
x
(
x
−
1
)
+
l
o
g
|
√
x
+
√
x
−
1
|
+
c
∫
x
√
x
−
1
d
x
=
√
x
(
x
−
1
)
+
l
o
g
|
√
x
+
√
x
−
1
|
+
c
⋅
∫
x
e
2
x
(
1
+
x
)
d
x
equal to
Report Question
0%
x
e
x
2
+
c
0%
(
e
x
)
2
2
r
0%
(
1
+
x
)
2
2
+
c
0%
(
x
e
x
)
2
2
Explanation
I
=
∫
x
e
2
x
(
1
+
x
)
d
x
=
∫
x
e
2
x
d
x
+
∫
x
2
e
2
x
d
x
=
e
2
x
x
2
2
−
∫
x
2
e
2
x
d
x
+
∫
x
2
e
2
x
d
x
I
=
(
x
e
x
)
2
2
The value of
∫
x
2
2
3
x
d
x
=
Report Question
0%
x
2
2
3
x
4
log
2
−
x
.2
3
x
+
1
9.
(
log
2
)
3
+
2
3
x
+
1
27
(
log
2
)
+
c
0%
x
2
2
3
x
3
log
2
−
x
.2
3
x
+
1
9.
(
log
2
)
+
2
3
x
+
1
27
(
log
2
)
+
c
0%
x
2
2
3
x
3
log
2
−
2
3
x
+
1
.
x
3
2
(
log
2
)
2
+
2
3
x
+
1
3
3
(
log
2
)
3
+
c
0%
x
2
2
3
x
log
2
−
2
3
x
+
1
.
x
3
2
(
log
2
)
2
+
2
3
x
+
1
(
log
2
)
3
+
c
Explanation
∫
x
2
⋅
2
3
x
d
x
⋅
=
x
2
∫
2
3
x
d
x
+
∫
2
x
∫
2
3
x
⋅
d
x
⋅
=
x
2
2
3
x
l
o
g
8
−
∫
2
x
2
3
x
l
o
g
8
d
x
=
x
2
2
3
x
l
o
g
8
−
2
l
o
g
8
[
x
2
3
x
.
l
o
g
8
−
∫
2
3
x
l
o
g
8
.
d
x
]
=
x
2
2
3
x
l
o
g
8
−
2
l
o
g
8
[
x
2
3
x
l
o
g
8
−
1
l
o
g
8
2
3
x
l
o
g
8
]
+
c
.
=
x
2
2
3
x
l
o
g
8
−
2
3
x
+
1
⋅
x
3
2
(
l
o
g
2
)
2
+
2
3
x
+
1
3
3
(
l
o
g
2
)
3
+
c
∫
log
x
d
x
=
Report Question
0%
x
(
log
x
−
x
)
+
c
0%
x
(
log
x
)
+
c
0%
x
(
log
x
−
1
)
+
c
0%
x
(
log
x
+
x
)
+
c
Explanation
∫
l
o
g
x
d
x
=
∫
l
o
g
x
−
1
d
x
=
l
o
g
x
∫
1
−
d
x
−
∫
(
1
/
x
.
√
1.
d
x
)
d
x
=
l
o
g
x
(
x
)
−
x
+
c
=
x
(
l
o
g
x
−
1
)
+
c
∵
∫
l
o
g
x
d
x
=
x
(
l
o
g
x
−
1
)
+
c
Evaluate :
∫
cot
2
x
(
c
o
sec
2
x
+
c
o
sec
x
)
d
x
Report Question
0%
x
−
sin
x
+
C
0%
x
+
cos
x
+
C
0%
sin
x
−
x
+
C
0%
2
tan
(
a
x
2
)
+
c
Explanation
∫
cot
2
x
(
c
o
sec
2
x
+
c
o
sec
x
)
d
x
∫
c
o
s
e
c
2
x
−
1
c
o
s
e
c
x
(
c
o
s
e
c
x
+
1
)
d
x
∫
(
c
o
s
e
c
x
−
1
)
(
c
o
s
e
c
x
+
1
)
c
o
s
e
c
x
(
c
o
s
e
c
x
+
1
)
d
x
∫
c
o
s
e
c
x
−
1
c
o
s
e
c
x
d
x
=
∫
(
1
−
s
i
n
x
)
d
x
=
x
+
c
o
s
x
+
c
∵
−
∫
s
i
n
x
.
d
x
=
c
o
s
x
+
c
∫
[
c
o
s
x
x
−
sin
x
log
x
]
d
x
=
Report Question
0%
(
log
x
)
s
i
n
x
+
c
0%
(
log
x
)
(
c
o
s
x
)
+
c
0%
2
(
log
x
)
(
c
o
s
x
)
+
c
0%
−
(
log
x
)
s
i
n
x
+
c
Explanation
∫
[
c
o
s
x
x
−
s
i
n
x
l
o
g
x
]
d
x
∫
c
o
s
x
x
d
x
−
∫
s
i
n
x
l
o
g
x
d
x
c
o
s
x
∫
1
x
d
x
+
∫
s
i
n
x
l
o
g
x
.
d
x
−
∫
s
i
n
x
l
o
g
x
d
x
.
=
(
c
o
s
x
)
(
l
o
g
x
)
+
c
Hence,
∫
[
c
o
s
x
x
−
(
s
i
n
x
)
(
l
o
g
x
)
]
d
x
=
(
c
o
s
x
)
(
l
o
g
x
)
+
c
Evaluate
∫
x
2
e
x
d
x
=
Report Question
0%
e
x
(
x
2
−
2
x
+
2
)
+
c
0%
e
x
(
x
2
+
2
x
+
2
)
+
c
0%
x
2
+
e
x
+
c
0%
e
x
(
x
2
+
x
+
2
)
+
c
Explanation
Let
I
=
∫
x
2
e
x
d
x
=
x
2
∫
e
x
d
x
−
∫
(
d
x
2
d
x
⋅
∫
e
x
d
x
)
.
d
x
=
x
2
e
x
−
∫
(
2
x
e
x
)
d
x
⋅
=
x
2
e
x
−
2
∫
x
e
x
⋅
d
x
⋅
Taking
∫
x
e
x
=
x
e
x
−
e
x
⋅
....(by using by parts)
Therefore,
I
=
x
2
e
x
−
2
[
x
e
x
−
e
x
]
+
c
=
e
x
[
x
2
−
2
x
+
2
]
+
c
∫
e
log
x
cos
x
d
x
=
Report Question
0%
x
s
i
n
x
−
c
o
s
x
+
c
0%
x
2
s
i
n
x
+
c
o
s
2
x
+
c
0%
x
s
i
n
x
+
c
o
s
x
+
c
0%
x
s
i
n
x
+
s
i
n
x
+
c
Explanation
∫
e
l
o
g
x
c
o
s
x
d
x
=
∫
x
c
o
s
x
d
x
.
=
x
∫
c
o
s
x
d
x
−
∫
(
1
∫
c
o
s
x
d
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
.
∫
x
cos
x
log
x
−
sin
x
x
(
log
x
)
2
d
x
is equal to
Report Question
0%
sin
x
log
x
+
c
0%
log
x
+
sin
x
+
c
0%
log
x
sin
x
+
c
0%
cos
x
log
x
+
c
Explanation
∫
x
cos
x
log
x
−
sin
x
x
(
log
x
)
2
d
x
=
∫
cos
x
log
x
d
x
−
∫
sin
x
x
(
log
x
)
2
d
x
Using Integration by parts in the first expression, we get
=
1
log
x
∫
cos
x
d
x
−
∫
(
d
(
1
log
x
)
d
x
⋅
∫
cos
x
d
x
)
d
x
−
∫
sin
x
x
(
log
x
)
2
d
x
=
1
log
x
(
sin
x
)
+
∫
1
x
(
log
x
)
2
sin
x
d
x
−
∫
sin
x
x
(
log
x
)
2
d
x
=
sin
x
log
x
+
c
where,
c
is a constant of integration.
The value of
∫
x
(
c
o
s
e
c
x
c
o
t
x
)
d
x
=
is
Report Question
0%
x
c
o
s
e
c
x
−
l
o
g
|
t
a
n
x
/
2
|
+
c
0%
2
−
x
c
o
s
e
c
x
+
log
|
t
a
n
x
2
|
+
c
0%
x
c
o
s
e
c
x
−
2
log
|
t
a
n
x
/
2
|
+
c
0%
x
c
o
t
x
−
l
o
g
|
t
a
n
x
2
|
+
c
Explanation
∫
x
(
c
o
s
e
c
x
c
o
t
)
d
x
=
x
∫
(
c
o
s
e
c
x
c
o
t
x
)
d
x
−
∫
1
(
∫
c
o
s
e
c
x
c
o
t
x
d
x
)
d
x
=
x
[
−
c
o
s
e
c
x
]
+
c
+
∫
(
c
o
s
e
c
x
)
d
x
=
−
x
c
o
s
e
c
x
+
c
+
l
o
g
|
c
o
s
e
c
x
−
c
o
t
x
|
=
−
x
c
o
s
e
c
x
+
c
+
l
o
g
|
1
−
c
o
s
x
s
i
n
x
|
=
2
−
x
c
o
s
e
c
x
+
c
1
+
l
o
g
|
t
a
n
x
/
2
|
∫
x
c
o
s
e
c
x
c
o
t
x
d
x
=
2
−
x
c
o
s
e
c
x
+
c
1
+
l
o
g
|
t
a
n
x
2
|
Solve
∫
tan
−
1
x
d
x
Report Question
0%
x
tan
x
−
log
|
1
+
x
2
|
+
c
0%
x
tan
−
1
x
−
1
2
log
|
1
+
x
2
|
+
c
0%
x
tan
−
1
x
+
log
√
1
+
x
2
+
c
0%
x
tan
x
+
log
|
1
+
x
2
|
+
c
Explanation
∫
tan
−
1
x
.1
d
x
Let us assume here
tan
−
1
x
is the first function and constant 1 is the second function. Then the integral of the second function is x.
∫
u
.
v
d
x
=
u
∫
v
d
x
−
∫
(
d
u
d
x
∫
v
d
x
)
d
x
Now by integrating the functions by parts, we get
=
tan
−
1
x
∫
1
d
x
−
∫
(
d
t
a
n
−
1
x
d
x
∫
1
d
x
)
d
x
=
x
tan
−
1
x
−
∫
1
1
+
x
2
x
d
x
=
x
tan
−
1
x
+
c
−
1
2
∫
2
x
d
x
1
+
x
2
=
−
x
tan
−
1
x
−
1
2
log
(
1
+
x
2
)
+
c
=
∫
tan
−
1
x
d
x
=
x
tan
−
1
x
−
1
2
log
(
1
+
x
2
)
+
c
Evaluate
∫
log
(
x
/
e
)
(
log
x
)
2
d
x
Report Question
0%
log
x
x
+
c
0%
x
log
x
+
c
0%
x
l
o
g
(
x
)
2
+
c
0%
(
log
x
)
2
x
+
c
Explanation
∫
log
(
x
/
e
)
(
log
x
)
2
d
x
=
∫
log
x
−
1
(
log
x
)
2
=
∫
[
1
(
log
x
)
−
1
(
log
x
)
2
]
d
x
=
∫
1
(
log
x
)
d
x
−
∫
1
(
log
x
)
2
d
x
=
1
(
log
x
)
∫
1.
d
x
+
∫
1
(
log
x
)
2
d
x
−
∫
1
(
log
x
)
2
d
x
..... [Using integration by parts]
=
x
log
x
+
c
∴
∫
log
(
x
/
e
)
(
log
x
)
2
d
x
=
x
log
x
+
c
∫
x
2
c
o
s
h
4
x
d
x
=
Report Question
0%
x
2
4
s
i
n
h
4
x
−
x
8
c
o
s
h
4
x
+
sinh
4
x
32
+
c
0%
x
2
4
s
i
n
h
4
x
+
x
8
cosh
4
x
−
sinh
4
x
32
+
c
0%
x
2
4
s
i
n
h
4
x
−
x
8
cosh
4
x
−
sinh
4
x
32
+
c
0%
x
2
4
s
i
n
h
4
x
+
x
8
cosh
4
x
−
sin
h
4
x
64
+
Explanation
∫
x
2
c
o
s
h
4
x
.
d
x
x
2
∫
c
o
s
h
4
x
d
x
−
∫
2
x
[
(
∫
c
o
s
h
4
x
)
d
x
]
d
x
=
x
2
4
s
i
n
h
4
x
−
1
/
2
∫
(
x
s
i
n
h
4
x
)
d
x
(
1
)
∫
x
s
i
n
h
4
x
d
x
=
x
4
c
o
s
h
4
x
−
∫
c
o
s
h
4
4
x
d
x
=
(
x
4
c
o
s
h
4
x
−
s
i
n
16
4
x
)
−
(
2
)
Using (2) in (1),
=
x
2
4
s
i
n
h
4
x
−
x
8
c
o
s
h
4
x
+
s
i
n
h
4
x
32
+
c
∫
x
2
c
o
s
h
4
x
d
x
=
x
2
4
s
i
n
h
4
x
−
x
/
8
c
o
s
h
4
x
+
s
i
n
h
4
x
32
+
c
∫
c
o
t
−
1
(
x
−
1
x
+
1
)
d
x
=
Report Question
0%
π
4
x
+
x
cot
−
1
x
−
1
2
log
(
1
+
x
2
)
+
c
0%
x
cot
−
1
x
+
1
2
log
(
1
+
x
2
)
+
c
0%
π
4
x
−
1
2
log
(
1
+
x
2
)
+
c
0%
π
4
x
+
x
cot
−
1
x
+
1
2
log
(
1
+
x
2
)
+
c
Explanation
c
o
s
e
c
θ
=
√
1
+
x
2
s
i
n
θ
=
1
√
1
+
x
2
∫
c
o
t
−
1
(
x
−
1
x
+
1
)
d
x
x
=
c
o
t
θ
∫
c
o
t
−
1
(
c
o
t
θ
−
1
c
o
t
θ
+
1
)
[
c
o
s
e
c
2
θ
]
d
θ
−
∫
c
o
t
−
1
[
c
o
t
(
π
/
4
+
θ
)
]
(
c
o
s
e
c
2
θ
)
d
θ
=
−
∫
(
π
/
4
+
θ
(
c
o
s
e
c
2
θ
)
)
d
θ
=
−
[
∫
π
/
4
c
o
s
e
c
γ
θ
d
θ
+
∫
θ
c
o
s
e
c
γ
θ
d
θ
]
=
−
[
−
π
/
4
c
o
t
θ
+
[
θ
(
−
c
o
t
θ
)
+
∫
c
o
t
θ
d
θ
]
=
π
/
4
c
o
t
θ
+
c
o
t
θ
−
l
o
g
|
s
i
n
θ
|
+
c
=
π
/
4
x
+
x
c
o
t
−
1
x
−
1
/
2
l
o
g
(
√
1
+
x
2
)
+
c
∫
c
o
t
−
1
(
x
−
1
x
+
1
)
d
x
=
π
/
4
x
+
x
c
o
t
−
1
x
−
1
/
2
l
o
g
(
√
1
+
x
2
)
+
c
∫
e
x
c
o
s
e
c
x
(
1
−
cot
x
)
d
x
=
Report Question
0%
e
x
cot
x
+
c
0%
e
x
c
o
s
e
c
x
cot
x
+
c
0%
e
x
c
o
s
e
c
x
+
c
0%
−
e
x
cot
x
+
c
Explanation
∫
e
x
c
o
s
e
c
x
(
1
−
c
o
t
x
)
d
x
∫
e
x
(
c
o
s
e
c
x
−
c
o
s
e
x
c
o
t
x
)
d
x
∫
e
x
c
o
s
e
c
x
d
x
−
∫
e
x
(
c
o
s
e
c
x
.
c
o
t
x
)
d
x
∫
e
x
(
c
o
s
e
c
x
c
o
t
x
)
d
x
=
e
x
∫
c
o
s
e
c
x
c
o
t
x
d
x
+
∫
e
x
c
o
s
e
c
x
.
d
x
=
e
x
c
o
s
e
c
x
+
∫
c
o
s
e
c
x
d
x
.
∫
e
x
c
o
s
e
c
x
d
x
−
(
−
e
x
c
o
s
e
c
x
+
∫
c
o
s
e
c
x
d
x
)
+
c
=
∫
e
x
c
o
s
e
c
x
d
x
+
e
x
c
o
s
e
c
x
−
∫
e
x
c
o
s
e
c
x
d
x
=
e
x
c
o
s
e
c
x
+
c
∫
x
2
a
x
d
x
=
Report Question
0%
a
x
[
x
2
log
a
−
2
x
(
log
a
)
2
+
2
(
log
a
)
3
]
+
c
0%
a
x
[
x
2
log
a
+
2
x
(
log
a
)
2
−
2
(
log
a
)
3
]
+
c
0%
a
x
[
x
2
log
a
+
2
x
(
log
a
)
2
+
2
(
log
a
)
3
]
+
c
0%
a
x
(
x
log
a
−
1
(
log
a
)
2
)
+
c
Explanation
∫
x
2
a
x
d
x
=
x
2
∫
a
x
d
x
−
∫
[
d
x
2
d
x
∫
a
x
d
x
]
d
x
x
2
a
x
l
o
g
a
−
∫
[
2
x
∫
a
x
d
x
]
d
x
x
2
a
x
l
o
g
a
−
2
∫
x
a
x
l
o
g
a
d
x
x
2
a
x
l
o
g
a
−
2
l
o
g
a
∫
[
x
a
x
]
d
x
∫
x
a
x
d
x
=
x
a
a
a
−
∫
1
a
x
(
l
o
g
a
)
d
x
=
x
a
x
l
o
g
a
−
a
x
(
l
o
g
a
)
2
=
x
2
a
x
l
o
g
a
−
2
l
o
g
a
[
x
a
x
l
o
g
a
−
a
x
(
l
o
g
a
)
2
]
=
x
2
a
x
l
o
g
a
−
2
l
o
g
a
[
x
a
x
l
o
g
a
−
a
x
(
l
o
g
a
)
2
]
a
x
[
x
2
l
o
g
a
−
2
x
(
l
o
g
a
)
2
+
2
(
l
o
g
a
)
3
]
+
c
lf polynomials
P
and
Q
satisfy
∫
(
(
3
x
−
1
)
cos
x
+
(
1
−
2
x
)
sin
x
)
d
v
=
P
cos
x
+
Q
sin
x
+
R
(ignoring the constant of integration) then
Report Question
0%
P
=
3
x
−
2
0%
Q
=
2
+
x
0%
P
=
3
(
x
−
1
)
0%
Q
=
3
(
x
−
1
)
Explanation
Let
I
=
∫
(
(
3
x
−
1
)
cos
x
+
(
1
−
2
x
)
sin
x
)
d
x
Using integration by parts, we evaluate the 2 expressions individually,
∫
(
3
x
−
1
)
cos
x
=
(
3
x
−
1
)
sin
x
−
∫
3
sin
x
d
x
=
(
3
x
−
1
)
sin
x
+
3
cos
x
+
c
Similarly,
∫
(
1
−
2
x
)
sin
x
d
x
=
−
(
1
−
2
x
)
cos
x
−
∫
(
−
2
)
(
−
cos
x
)
d
x
=
(
2
x
−
1
)
cos
x
−
2
sin
x
+
c
Adding both the expressions, we get
I
=
3
(
x
−
1
)
sin
x
+
2
(
x
+
1
)
cos
x
Ignoring the constant
Comparing it with the RHS, we get
P
=
2
(
x
+
1
)
and
Q
=
3
(
x
−
1
)
Hence, option 'D' is correct.
Evaluate
∫
arcsin
√
x
√
1
−
x
d
x
=
Report Question
0%
2
[
√
x
−
√
1
−
x
arc
sin
√
x
]
+
c
0%
2
[
√
x
+
√
1
−
x
arc
sin
√
x
]
+
c
0%
2
[
√
x
+
√
1
−
x
arc
cos
√
x
]
+
c
0%
2
[
√
x
−
√
1
−
x
arc
cos
√
x
]
+
c
Explanation
∫
sin
−
1
√
x
√
1
−
x
⋅
d
x
.
Put
x
=
sin
2
θ
⇒
d
x
=
2
sin
θ
cos
θ
d
θ
.
∴
∫
sin
−
1
√
x
√
1
−
x
⋅
d
x
.
=
∫
sin
−
1
(
sin
θ
)
√
1
−
sin
2
θ
⋅
d
θ
(
2
sin
θ
)
(
cos
θ
)
=
2
∫
θ
sin
θ
d
θ
=
2
[
θ
(
−
cos
θ
)
+
∫
cos
θ
d
θ
.
]
=
2
[
θ
(
−
cos
θ
)
+
sin
θ
]
+
c
=
2
[
sin
θ
−
θ
cos
θ
]
+
c
=
2
[
√
x
−
(
sin
−
1
√
x
)
√
1
−
x
]
+
c
∴
∫
sin
−
1
√
x
√
1
−
x
d
x
=
2
[
√
x
−
(
√
1
−
x
)
sin
−
1
√
x
]
+
c
∫
e
x
(
x
2
−
5
x
+
8
)
dx
=
e
x
f
(
x
)
+
c
then
f
(
x
)
Report Question
0%
x
2
−
5
x
+
12
0%
x
2
+
7
x
+
15
0%
x
2
−
7
x
−
15
0%
x
2
−
7
x
+
15
Explanation
∫
e
x
(
x
2
−
5
x
+
8
)
d
x
=
∫
e
x
x
2
d
x
−
5
∫
e
x
.
x
d
x
+
8
∫
e
x
d
x
=
x
2
e
x
=
2
∫
x
e
x
d
x
−
5
∫
e
x
x
+
8
∫
e
x
d
x
x
2
e
x
−
7
[
x
e
x
−
e
x
]
+
8
e
x
=
x
2
e
x
−
7
x
e
x
+
15
e
x
=
e
n
[
x
2
−
7
x
+
15
]
+
c
s
o
,
f
(
x
)
−
x
2
−
7
x
+
15
Evaluate
∫
x
(
sec
2
x
−
1
)
(
sec
2
x
+
1
)
d
x
Report Question
0%
x
tan
x
−
log
|
sec
x
|
+
x
2
2
+
c
0%
x
tan
x
−
log
|
sec
x
|
−
x
2
2
+
c
0%
x
tan
x
−
log
|
sec
x
2
|
+
c
0%
x
tan
x
−
log
|
sec
x
2
|
+
x
2
2
+
c
Explanation
∫
x
(
sec
2
x
−
1
sec
2
x
+
1
)
d
x
=
∫
x
(
1
−
cos
2
x
1
+
cos
2
x
)
d
x
=
∫
x
tan
2
x
d
x
∫
x
(
sec
2
x
−
1
)
d
x
=
∫
x
sec
2
x
d
x
−
∫
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, for the first part taking
x
as
u
and
s
e
c
2
x
as
v
.
=
x
∫
sec
2
x
d
x
−
∫
1
(
∫
sec
2
x
d
x
)
d
x
−
∫
x
d
x
=
x
tan
x
−
∫
tan
x
d
x
−
x
2
2
+
c
=
x
tan
x
−
log
|
sec
x
|
−
x
2
2
+
c
∫
x
(
sec
2
x
−
1
sec
2
x
+
1
)
d
x
=
x
tan
x
−
log
|
sec
x
|
−
x
2
2
+
c
∫
e
x
2
3
log
2
x
d
x
=
e
x
f
(
x
)
+
c
, then
f
(
x
)
=
Report Question
0%
x
3
−
3
x
2
+
6
x
−
6
0%
x
3
−
3
x
2
−
6
x
−
3
0%
x
3
−
3
x
2
+
6
x
+
6
0%
x
3
+
3
x
2
+
6
x
+
6
Explanation
Given,
∫
e
x
2
3
log
2
x
d
x
=
e
x
f
(
x
)
+
c
......(A)
Consider,
∫
e
x
2
3
log
2
x
d
x
=
∫
e
x
.
x
3
d
x
∫
e
x
×
x
3
d
x
=
x
3
∫
e
x
d
x
−
∫
3
x
2
e
x
d
x
=
x
3
e
x
−
3
∫
x
2
e
x
d
x
....(1)
∫
x
2
e
x
=
x
2
e
x
−
2
∫
x
e
x
d
x
.....(2)
∫
x
e
x
=
x
e
x
−
e
x
.....(3)
Using (3) in (2),
∫
x
2
e
x
=
x
2
e
x
−
2
(
x
e
x
−
e
x
)
....(4)
Using (4) in (1)
∫
e
x
.
x
3
d
x
=
x
3
e
x
−
3
[
x
2
e
x
−
2
(
x
e
x
−
e
x
)
]
+
c
=
x
3
e
x
−
3
x
2
e
x
+
6
x
e
x
−
6
e
x
+
c
=
e
x
[
x
3
−
3
x
2
+
6
x
−
6
]
+
c
On comparing with eqn (A), we get
f
(
x
)
=
x
3
−
3
x
2
+
6
x
−
6
∫
x
+
sin
x
1
+
cos
x
d
x
=
Report Question
0%
x
t
a
n
x
2
+
c
0%
x
c
o
t
x
2
+
c
0%
x
s
i
n
x
2
+
c
0%
x
c
o
s
x
2
+
c
Explanation
∫
x
+
sin
x
1
+
cos
x
d
x
=
1
2
∫
x
+
sin
x
cos
2
(
x
/
2
)
d
x
=
1
2
∫
x
sec
2
(
x
/
2
)
d
x
+
∫
tan
(
x
/
2
)
d
x
=
1
2
[
x
tan
(
x
/
2
)
1
/
2
−
∫
tan
(
x
/
2
)
1
/
2
d
x
]
+
∫
tan
(
x
/
2
)
d
x
+
C
=
x
tan
(
x
/
2
)
+
C
∫
cos
−
1
(
1
x
)
d
x
equal to
Report Question
0%
x
sec
−
1
x
+
cosh
−
1
x
+
c
0%
x
sec
−
1
x
−
cosh
−
1
x
+
c
0%
x
sec
−
1
x
−
sin
−
1
x
+
c
0%
x
sec
−
1
x
+
sin
−
1
x
+
c
Explanation
x
=
s
e
c
θ
d
x
=
s
e
c
θ
⋅
t
a
n
θ
d
θ
.
∫
c
o
x
−
1
(
c
o
s
θ
)
⋅
(
s
e
c
θ
⋅
t
a
n
θ
)
d
θ
.
∫
θ
(
s
e
c
θ
⋅
t
a
n
θ
)
d
θ
⋅
θ
∫
(
s
e
c
θ
⋅
t
a
n
θ
)
d
θ
−
∫
(
∫
(
s
e
c
θ
⋅
t
a
n
θ
)
d
θ
)
d
θ
∫
s
e
c
θ
t
a
n
θ
⋅
d
θ
=
s
e
c
θ
+
c
=
θ
s
e
c
θ
−
∫
s
e
c
θ
d
θ
=
θ
s
e
c
θ
+
(
l
o
g
∣
s
e
c
θ
+
t
a
n
θ
∣
)
+
1.
=
x
s
e
c
−
1
x
.
−
l
o
g
∣
x
+
√
x
2
−
1
∣
+
c
=
x
s
e
c
−
1
x
−
c
o
s
h
−
1
x
+
c
.
∫
c
o
s
−
1
(
1
x
)
=
x
s
e
c
−
1
x
−
c
o
s
h
−
1
x
+
c
.
Evaluate:
∫
log
10
x
d
x
Report Question
0%
(
x
−
1
)
log
e
x
+
c
0%
log
e
10.
x
log
e
(
x
e
)
+
c
0%
log
10
e
.
x
log
e
(
x
e
)
+
c
0%
1
x
+
c
Explanation
We know that,
∫
u
v
d
x
=
u
′
∫
v
d
x
−
∫
u
′
(
∫
v
d
x
)
d
x
+
C
(Integration by parts rule)
∴
we can write,
I
=
∫
log
10
x
d
x
=
log
10
e
∫
log
e
x
⋅
(
1
)
d
x
...as
log
10
x
=
log
e
x
log
e
10
=
log
10
e
⋅
log
e
x
Applying integration by parts;
u
=
log
e
x
v
=
1
u
′
=
1
x
∫
v
d
x
=
x
∴
I
=
log
10
e
(
(
log
e
x
)
x
−
∫
1
x
⋅
x
d
x
)
+
C
=
log
10
e
(
x
log
e
x
−
∫
d
x
)
+
C
I
=
log
10
e
⋅
x
(
log
e
x
−
1
)
+
C
=
log
10
e
⋅
x
(
log
e
x
−
log
e
e
)
+
C
∴
I
=
log
10
e
⋅
x
(
log
e
x
e
)
+
C
Evaluate
∫
x
e
x
(
x
+
1
)
2
d
x
Report Question
0%
e
x
[
1
x
+
1
]
+
c
0%
e
x
[
−
1
x
+
1
]
+
c
0%
e
x
x
+
c
0%
e
x
[
−
1
(
x
+
1
)
2
]
+
c
Explanation
I
=
∫
x
e
x
(
x
+
1
)
2
d
x
=
∫
(
x
+
1
−
1
)
e
x
(
x
+
1
)
2
d
x
=
∫
(
x
+
1
)
e
x
(
x
+
1
)
2
d
x
−
∫
e
x
(
x
+
1
)
2
d
x
=
∫
e
x
x
+
1
d
x
−
∫
e
x
(
x
+
1
)
2
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, for the first part taking
e
x
as
u
and
1
x
+
1
as
v
I
=
e
x
x
+
1
+
∫
e
x
(
x
+
1
)
2
d
x
−
∫
e
x
(
x
+
1
)
2
d
x
I
=
e
x
x
+
1
+
C
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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