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NDA Math Definite Integrals
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© examsnet.com
Question : 33
Total: 62
If
π
2
∫
0
d
x
3
cosx
+
5
=
k
cot
−
1
2
then what is the value of K?
[NDA II 2016]
1/4
1/2
1
2
Validate
Solution:
I
=
π
2
∫
0
d
x
3
cosx
+
5
I
=
π
2
∫
0
d
x
3
[
1
−
tan
2
x
2
1
+
tan
2
x
2
]
+
5
I
=
π
2
∫
0
(
1
+
tan
2
x
2
)
d
x
3
−
3
tan
2
x
2
+
5
+
5
tan
2
x
2
I
=
π
2
∫
0
sec
2
x
2
d
x
2
tan
2
x
2
+
8
I
=
1
2
π
2
∫
0
sec
2
x
2
d
x
tan
2
x
2
+
2
2
Put
tan
x
2
=
y
⇒
1
2
sec
2
x
2
d
x
=
d
y
⇒
I
=
1
∫
0
d
y
y
2
+
2
2
⇒
I
=
1
2
tan
−
1
(
y
2
)
⇒
I
=
1
2
tan
−
1
(
1
2
)
−
0
Also
I
=
1
2
tan
−
1
1
2
=
k
cot
−
1
(
2
)
(
∵
tan
−
1
(
x
)
=
cot
−
1
(
1
x
)
)
⇒
1
2
tan
−
1
(
1
2
)
=
k
tan
−
1
(
1
2
)
∴
k
=
1
2
© examsnet.com
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