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Test Index
JEE Main Math Class 12 Integrals Part 4 Questions
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© examsnet.com
Question : 5
Total: 51
The integral
π
4
∫
0
136
s
i
n
x
3
s
i
n
x
+
5
cos
x
dx
is equal to :
[5 Apr 2024 Shift 1]
3
π
−
50
log
e
2
+
20
log
e
5
3
π
−
25
log
e
2
+
10
log
e
5
3
π
−
10
log
e
(
2
√
2
)
+
10
log
e
5
3
π
−
30
log
e
2
+
20
log
e
5
Validate
Solution:
👈: Video Solution
I
=
π
∕
4
∫
0
136
s
i
n
x
3
s
i
n
x
+
5
cos
x
dx
136
s
i
n
x
=
A
(
3
s
i
n
x
+
5
cos
x
)
+
B
(
3
cos
x
−
5
s
i
n
x
)
136
=
3
A
−
5
B
. . . (1)
0
=
5
A
+
3
B
. . . (2)
3
B
=
−
5
A
⇒
B
=
−
5
3
A
136
=
3
A
−
5
(
−
5
3
A
)
136
=
3
A
+
25
3
A
136
=
34
A
3
⇒
A
=
136
×
3
34
=
12
B
=
−
5
3
(
12
)
=
−
20
I
=
π
∕
4
∫
0
A
(
3
s
i
n
x
+
5
cos
x
)
3
s
i
n
x
+
5
cos
x
+
π
∕
4
∫
0
B
(
3
cos
x
−
5
s
i
n
x
)
3
s
i
n
x
+
5
cos
x
=
A
(
x
)
0
π
∕
4
+
B
[
ln
(
3
s
i
n
x
+
5
cos
x
)
]
0
π
∕
4
=
12
(
π
4
)
−
20
ln
(
3
√
2
+
5
√
2
)
−
ln
(
0
+
5
)
=
3
π
−
20
ln
4
√
2
+
20
ln
5
=
3
π
−
20
×
5
2
ln
2
+
20
ln
5
=
3
π
−
50
ln
2
+
20
ln
5
© examsnet.com
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