Email
Chat with tutors
Login
Ask Questions, Get Answers
Menu
X
home
ask
tuition
questions
practice
papers
mobile
tutors
pricing
X
Recent questions in Integrals
Questions
>>
CBSE XII
>>
Math
>>
Integrals
∫
cos
x
cos
(
x
−
a
)
d
x
=
A
x
+
B
l
o
g
cos
(
x
−
α
)
,
t
h
e
n
(
a
)
A
=
cos
α
(
b
)
B
=
sin
α
(
c
)
A
=
sin
α
(
d
)
B
=
cos
α
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q25
p25
math
asked
Feb 1, 2013
by
meena.p
0
answers
If
∫
tan
4
x
d
x
=
a
tan
3
x
+
b
tan
x
+
c
x
,
t
h
e
n
(
a
)
a
=
1
3
(
b
)
b
=
−
1
(
c
)
a
=
1
(
d
)
c
=
1
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q24
p25
math
asked
Feb 1, 2013
by
meena.p
0
answers
If
∫
d
x
(
x
+
1
)
(
x
−
2
)
=
A
l
o
g
(
x
+
1
)
+
B
l
o
g
(
x
−
2
)
+
c
,
t
h
e
n
(
a
)
A
+
B
=
0
(
b
)
A
B
=
−
(
c
)
A
B
=
1
9
(
d
)
A
B
=
−
9
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q23
p25
math
asked
Feb 1, 2013
by
meena.p
0
answers
S
sin
2
x
l
o
g
cos
x
d
x
=
(
a
)
1
2
cos
2
x
−
c
o
s
2
x
l
o
g
cos
x
(
b
)
1
2
cos
2
x
+
cos
2
l
o
g
cos
x
(
c
)
cos
2
x
.
l
o
g
cos
x
(
d
)
cos
2
x
(
1
−
l
o
g
cos
x
)
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q22
p25
math
asked
Feb 1, 2013
by
meena.p
0
answers
∫
d
x
x
+
√
x
=
(
a
)
log
(
1
+
√
x
)
(
b
)
1
2
log
(
1
+
√
x
)
(
c
)
2
log
(
1
+
√
x
)
(
d
)
log
(
x
+
√
x
)
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q21
p25
math
asked
Feb 1, 2013
by
meena.p
0
answers
∫
d
x
(
2
−
x
)
√
1
+
x
=
(
a
)
2
tan
−
1
√
1
+
x
(
b
)
1
2
tan
−
1
√
1
+
x
(
c
)
tan
−
1
√
1
+
x
(
d
)
l
o
g
{
(
2
+
x
)
√
1
+
x
}
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q20
p25
math
asked
Feb 1, 2013
by
meena.p
0
answers
∫
e
l
o
g
(
1
+
1
x
2
)
d
x
x
2
+
1
x
2
=
(
a
)
1
√
2
tan
−
1
(
x
+
1
x
)
(
b
)
1
√
2
tan
−
1
(
x
−
1
x
√
2
)
(
c
)
√
2
tan
−
1
(
x
+
1
x
)
(
d
)
1
√
2
tan
−
1
(
1
x
−
x
)
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q19
p25
math
asked
Jan 31, 2013
by
meena.p
0
answers
1
x
(
x
4
−
1
)
4
=
(
a
)
log
[
x
4
x
4
−
1
]
(
b
)
1
2
log
[
x
2
+
1
x
2
+
1
]
(
c
)
1
4
l
o
g
[
x
4
−
1
x
4
]
(
d
)
log
x
(
x
2
−
1
x
2
+
1
]
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q18
p25
math
asked
Jan 31, 2013
by
meena.p
0
answers
3
x
d
x
√
1
−
9
x
=
(
a
)
log
e
3
sin
−
1
(
3
x
)
(
b
)
1
l
o
g
3
e
sin
−
1
(
3
x
)
(
c
)
log
e
3
sin
−
1
(
3
x
/
2
)
(
d
)
1
l
o
g
e
3
sin
−
1
(
3
x
/
2
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q17
p25
math
asked
Jan 31, 2013
by
meena.p
0
answers
∫
2
x
d
x
1
+
x
4
=
(
a
)
tan
−
1
(
x
2
)
(
b
)
1
2
tan
−
1
x
2
(
c
)
l
o
g
(
1
+
x
4
)
(
d
)
tan
−
1
(
1
x
2
)
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q16
p24
math
asked
Jan 31, 2013
by
meena.p
0
answers
∫
√
1
+
x
2
d
(
x
2
)
=
(
a
)
2
3
x
(
1
+
x
2
)
3
/
2
(
b
)
2
3
(
1
+
x
2
)
3
/
2
(
c
)
2
x
3
(
1
+
x
2
)
3
/
2
(
d
)
2
x
3
(
1
+
x
2
)
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q15
p24
math
asked
Jan 31, 2013
by
meena.p
0
answers
∫
d
x
x
−
X
3
=
(
a
)
1
2
l
o
g
x
2
1
−
x
2
(
b
)
1
2
l
o
g
1
−
x
2
x
2
(
c
)
l
o
g
(
1
−
x
)
x
(
1
+
x
)
(
d
)
l
o
g
x
(
1
−
x
2
)
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q14
p24
math
asked
Jan 31, 2013
by
meena.p
0
answers
∫
d
x
x
√
1
−
x
3
=
(
a
)
1
3
l
o
g
|
√
1
−
x
2
+
1
√
1
−
x
2
−
1
|
(
b
)
1
3
l
o
g
|
√
1
−
x
2
−
1
√
1
−
x
2
+
1
|
(
c
)
2
3
l
o
g
|
1
√
1
−
x
3
|
(
d
)
1
3
l
o
g
|
1
−
x
3
|
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q13
p24
math
asked
Jan 31, 2013
by
meena.p
0
answers
∫
sin
2
x
−
cos
2
x
sin
2
x
−
cos
2
x
d
x
=
(
a
)
tan
x
+
cot
x
(
b
)
tan
x
−
cot
x
(
c
)
tan
x
+
sec
x
(
d
)
tan
x
+
c
o
s
e
c
x
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q12
p24
math
asked
Jan 31, 2013
by
meena.p
0
answers
∫
x
+
sin
x
1
+
cos
x
d
x
=
(
a
)
tan
x
2
(
b
)
x
tan
x
2
(
c
)
cot
x
2
(
d
)
x
cot
x
2
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q11
p24
math
asked
Jan 31, 2013
by
meena.p
0
answers
∫
x
tan
−
1
x
d
x
=
(
a
)
(
x
2
+
1
2
)
tan
−
1
x
−
x
2
(
b
)
(
x
2
+
1
2
)
+
tan
−
1
x
−
x
(
c
)
(
x
2
+
1
)
tan
−
1
x
−
x
(
d
)
(
x
2
+
1
)
tan
−
1
x
+
x
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q10
p24
math
asked
Jan 30, 2013
by
meena.p
0
answers
∫
x
sin
−
1
x
√
1
−
x
2
d
x
=
(
a
)
√
1
−
x
2
sin
−
1
x
(
b
)
x
sin
−
1
x
(
c
)
x
−
√
1
−
x
2
sin
−
1
x
(
d
)
(
sin
−
1
x
)
2
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q9
p24
math
asked
Jan 30, 2013
by
meena.p
0
answers
∫
x
3
e
x
2
d
x
=
(
a
)
x
2
(
e
x
2
−
1
)
(
b
)
1
2
x
2
(
e
x
2
−
1
)
(
c
)
1
2
e
x
2
(
x
2
−
1
)
(
d
)
1
2
(
e
x
−
1
)
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q8
p23
math
asked
Jan 30, 2013
by
meena.p
0
answers
∫
√
x
√
a
3
−
x
3
d
x
=
(
a
)
2
3
(
x
a
)
3
/
2
(
b
)
2
3
sin
−
1
(
x
a
)
3
/
2
(
c
)
2
3
cos
−
1
(
x
a
)
3
/
2
(
d
)
2
3
sin
−
1
(
a
x
)
3
/
2
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q7
p23
math
asked
Jan 30, 2013
by
meena.p
0
answers
∫
e
x
d
x
e
2
x
+
1
=
(
a
)
l
o
g
(
e
x
+
e
−
x
)
(
b
)
l
o
g
(
e
2
x
+
1
)
(
c
)
tan
−
1
(
e
x
)
(
d
)
tan
−
1
(
2
e
x
)
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q6
p23
math
asked
Jan 30, 2013
by
meena.p
0
answers
∫
sin
2
x
d
x
a
2
cos
2
x
+
b
2
sin
2
x
sin
2
x
=
(
a
)
(
b
−
a
)
l
o
g
(
a
2
cos
2
x
+
b
2
sin
2
x
)
(
b
)
1
b
−
a
l
o
g
(
a
2
cos
2
x
+
b
2
sin
2
x
)
(
c
)
1
b
2
−
a
2
l
o
g
(
a
2
cos
2
x
+
b
2
sin
2
x
)
(
d
)
1
a
2
+
b
2
l
o
g
(
a
2
cos
2
x
+
b
2
sin
2
x
)
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q5
p23
math
asked
Jan 30, 2013
by
meena.p
0
answers
∫
d
x
sin
x
+
√
3
cos
x
=
(
a
)
1
2
l
o
g
tan
[
π
2
+
π
6
]
(
b
)
1
2
l
o
g
{
c
o
s
e
c
(
x
+
π
3
)
−
cot
(
x
+
π
3
)
}
(
c
)
1
2
l
o
g
{
sec
(
x
−
π
6
)
+
tan
(
x
−
π
6
)
}
(
d
)
−
1
2
l
o
g
{
c
o
s
e
c
(
x
+
π
3
)
+
cot
(
x
+
π
3
)
}
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q4
p23
math
asked
Jan 30, 2013
by
meena.p
0
answers
∫
√
tan
x
sin
x
cos
x
d
x
=
(
a
)
2
√
tan
x
(
b
)
2
√
cot
x
(
c
)
√
cot
x
(
d
)
√
tan
x
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q3
p23
math
asked
Jan 30, 2013
by
meena.p
0
answers
∫
tan
(
l
o
g
)
x
x
d
x
=
(
a
)
l
o
g
cos
(
l
o
g
x
)
(
b
)
l
o
g
sec
(
l
o
g
x
)
(
c
)
l
o
g
sin
(
l
o
g
x
)
(
d
)
−
l
o
g
cos
(
l
o
g
x
)
cbse
class12
kvquestionbank2012
additionalproblem
ch7
q2
p23
math
asked
Jan 30, 2013
by
meena.p
0
answers
choose the correct alternative
∫
cos
√
x
√
x
d
x
=
(
a
)
sin
√
x
(
b
)
2
cos
√
x
(
c
)
2
sin
√
x
(
d
)
√
cos
x
x
cbse
class12
additionalproblem
kvquestionbank2012
ch7
q1
p23
math
asked
Jan 30, 2013
by
meena.p
0
answers
The value of
π
∫
−
π
sin
3
x
cos
2
x
d
x
is___________.
cbse
class12
ch7
q63
p169
fitb
exemplar
sec-a
easy
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
∫
sin
x
3
+
4
cos
2
x
d
x
=___________.
cbse
class12
ch7
q62
p169
fitb
exemplar
sec-b
difficult
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
If
∫
1
1
+
4
x
2
d
x
=
π
8
,
then a=____________.
cbse
class12
ch7
q61
p169
fitb
exemplar
sec-b
easy
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
∫
x
+
3
(
x
+
4
)
2
e
x
d
x
=___________.
cbse
class12
ch7
q60
p169
fitb
exemplar
sec-b
easy
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
π
2
∫
0
cos
x
e
sin
x
d
x
is equal to ____________.
cbse
class12
ch7
q59
p169
fitb
exemplar
sec-b
medium
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
π
2
∫
0
√
1
−
sin
2
x
d
x
is equal to
(
A
)
0
(
B
)
2
(
C
)
1
(
D
)
−
1
cbse
class12
ch7
q58
p169
objective
exemplar
sec-b
easy
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
π
4
∫
−
π
4
d
x
1
+
cos
2
x
is equal to
(
A
)
1
(
B
)
2
(
C
)
3
(
D
)
4
cbse
class12
ch7
q57
p169
objective
exemplar
sec-b
easy
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
If
x
3
d
x
√
1
−
x
2
=
a
(
1
−
x
2
)
3
2
+
b
√
1
−
x
2
+
C
,then
cbse
class12
ch7
q56
p168
objective
exemplar
sec-b
easy
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
∫
x
+
sin
x
1
+
cos
x
d
x
is equal to
(
A
)
log
∣
1
+
cos
x
∣
+
C
(
B
)
log
∣
x
+
sin
x
∣
+
C
(
C
)
x
−
tan
x
2
+
C
(
D
)
x
tan
x
2
+
C
cbse
class12
ch7
q55
p168
objective
exemplar
sec-b
medium
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
∫
x
3
x
+
1
is equal to
(
A
)
x
+
x
2
2
+
x
3
3
−
l
o
g
∣
1
−
x
∣
+
C
(
B
)
x
+
x
2
2
−
x
3
3
−
l
o
g
∣
1
−
x
∣
+
C
(
C
)
x
−
x
2
2
−
x
3
3
−
l
o
g
∣
1
+
x
∣
+
C
(
D
)
x
−
x
2
2
+
x
3
3
−
l
o
g
∣
1
+
x
∣
+
C
cbse
class12
ch7
q54
p168
objective
exemplar
sec-b
easy
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
If
∫
d
x
(
x
+
2
)
(
x
2
+
1
)
=
a
log
∣
1
+
x
2
∣
+
b
tan
−
1
x
+
1
5
log
∣
x
+
2
∣
+
C
,then
(
A
)
a
=
−
1
10
,
b
=
−
2
5
(
B
)
a
=
1
10
,
b
=
−
2
5
(
C
)
a
=
−
1
10
,
b
=
2
5
(
D
)
a
=
1
10
,
b
=
2
5
cbse
class12
ch7
q53
p168
objective
exemplar
sec-c
difficult
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
∫
x
9
(
4
x
2
+
1
)
6
d
x
is equal to
(
A
)
1
5
x
(
4
+
1
x
2
)
−
5
+
C
(
B
)
1
5
(
4
+
1
x
2
)
−
5
+
C
(
C
)
1
10
x
(
1
+
4
)
−
5
+
C
(
D
)
1
10
(
1
x
2
+
4
)
−
5
+
C
cbse
class12
ch7
q52
p167
objective
exemplar
sec-b
medium
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
∫
e
x
(
1
−
x
1
+
x
2
)
2
d
x
is equal to
cbse
class12
ch7
q51
p167
objective
exemplar
sec-a
medium
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
∫
tan
−
1
√
x
is equal to which of the following options:
cbse
class12
ch7
q50
p166
objective
exemplar
sec-b
medium
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
∫
d
x
sin
(
x
−
a
)
sin
(
x
−
b
)
is equal to
(
A
)
sin
(
b
−
a
)
log
|
sin
(
x
−
b
)
sin
(
x
−
a
)
|
+
C
(
B
)
c
o
s
e
c
(
b
−
a
)
log
|
sin
(
x
−
a
)
sin
(
x
−
b
)
|
+
C
(
C
)
c
o
s
e
c
(
b
−
a
)
log
|
sin
(
x
−
b
)
sin
(
x
−
a
)
|
+
C
(
D
)
sin
(
b
−
a
)
log
|
sin
(
x
−
a
)
sin
(
x
−
b
)
|
+
C
cbse
class12
ch7
q49
p167
objective
exemplar
sec-b
difficult
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
∫
cos
2
x
−
cos
2
θ
cos
x
−
cos
θ
d
x
is equal to
(
A
)
2
(
sin
x
+
x
cos
θ
)
+
C
(
B
)
2
(
sin
x
−
x
cos
θ
)
+
C
(
C
)
2
(
sin
x
+
2
x
cos
θ
)
+
C
(
D
)
2
(
sin
x
−
2
x
cos
θ
)
+
C
cbse
class12
ch7
q48
p166
objective
exemplar
sec-b
easy
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
Evaluate the following:
π
4
∫
−
π
4
log
(
sin
x
+
cos
x
)
d
x
cbse
class12
ch7
q47
p166
exemplar
sec-c
difficult
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
Evaluate the following:
π
∫
0
x
log
sin
x
d
x
cbse
class12
ch7
q46
p166
exemplar
sec-b
difficult
math
asked
Jan 15, 2013
by
sreemathi.v
1
answer
Evaluate the following:
1
∫
0
x
l
o
g
(
1
+
2
x
)
d
x
cbse
class12
ch7
q45
p166
exemplar
sec-b
medium
math
asked
Jan 14, 2013
by
sreemathi.v
1
answer
Evaluate the following:
π
2
∫
0
d
x
(
a
2
cos
2
x
+
b
2
sin
2
x
)
2
(Hint:Divide Numerator and Denominator by
cos
4
x
)
cbse
class12
ch7
q44
p166
exemplar
sec-c
difficult
math
asked
Jan 14, 2013
by
sreemathi.v
1
answer
Evaluate :
∫
√
tan
x
d
x
cbse
class12
ch7
q43
p166
exemplar
sec-c
difficult
math
asked
Jan 14, 2013
by
sreemathi.v
1
answer
Evaluate the following:
∫
e
−
3
x
cos
3
x
d
x
cbse
class12
ch7
q42
p166
exemplar
sec-c
difficult
math
asked
Jan 14, 2013
by
sreemathi.v
1
answer
Evaluate the following:
π
2
∫
π
3
√
1
+
cos
x
(
1
−
cos
x
)
5
2
cbse
class12
ch7
q41
p166
exemplar
sec-b
difficult
math
asked
Jan 14, 2013
by
sreemathi.v
1
answer
Evaluate the following:
∫
sin
−
1
√
x
a
+
x
d
x
(
H
i
n
t
:
P
u
t
x
=
a
tan
2
θ
)
cbse
class12
ch7
q40
p166
exemplar
sec-c
medium
math
asked
Jan 14, 2013
by
sreemathi.v
1
answer
Evaluate the following:
∫
e
tan
−
1
x
(
1
+
x
+
x
2
1
+
x
2
)
d
x
cbse
class12
ch7
q39
p166
exemplar
sec-b
difficult
math
asked
Jan 14, 2013
by
sreemathi.v
1
answer
Page:
« prev
1
2
3
4
5
6
...
10
next »
Home
Ask
Tuition
Questions
Practice
Your payment for
is successful.
Continue
...
Please Wait