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KEAM 2016 Math Paper
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© examsnet.com
Question : 111
Total: 120
∫
1
8
sin
2
x
+
1
dx
is equal to
sin
−
1
(
tan
x
)
+
C
1
3
sin
−
1
(
tan
x
)
+
C
1
3
tan
−
1
(
3
tan
x
)
+
C
tan
−
1
(
3
tan
x
)
+
C
sin
−
1
(
3
tan
x
)
+
C
Validate
Solution:
Let
I
=
∫
1
8
sin
2
x
+
1
=
∫
sec
2
x
8
tan
2
x
+
sec
2
x
d
x
[dividing numerator and denominator by
cos
2
x
]
=
∫
sec
2
x
1
+
(
3
tan
x
)
2
d
x
Let
t
=
3
tan
x
⇒
d
t
d
x
=
3
sec
2
x
⇒
d
t
3
=
sec
2
x
d
x
∴
I
=
1
3
∫
d
t
1
+
(
t
)
2
=
1
3
tan
−
1
(
t
)
+
C
=
1
3
tan
−
1
(
3
tan
x
)
+
C
© examsnet.com
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