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TS EAMCET 2017 Code A Solved Paper
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© examsnet.com
Question : 6
Total: 160
π
∫
0
x
dx
4
c
o
s
2
x
+
9
s
i
n
2
x
=
π
2
12
π
2
4
π
2
6
π
2
3
Validate
Solution:
Consider the integration,
I
=
π
∫
0
x
d
x
4
cos
2
x
+
9
sin
2
x
.
.
.
.
.
(
1
)
The above integration can be simplified as,
I
=
π
∫
0
x
d
x
4
cos
2
x
+
9
sin
2
x
l
=
π
∫
0
(
π
−
x
)
d
x
4
cos
2
(
π
−
x
)
+
9
sin
2
(
π
−
x
)
=
π
∫
0
(
π
−
x
)
d
x
4
cos
2
x
+
9
sin
2
x
.
.
.
.
(
2
)
Add equation (1) and (2).
2
I
=
π
∫
0
π
d
x
4
cos
2
x
+
9
sin
2
x
=
π
∫
0
π
sec
2
x
d
x
4
+
9
tan
2
x
=
∫
0
π
2
2
π
sec
2
x
d
x
4
+
9
tan
2
x
I
=
π
9
∫
0
π
2
sec
2
x
d
x
4
9
+
tan
2
x
Put
tan
x
=
t
sec
2
x
d
x
=
d
t
At
x
=
0
,
t
=
0
x
=
π
2
,
t
=
∞
Then
I
=
π
9
∞
∫
0
d
t
(
2
3
)
2
+
t
2
=
π
9
×
3
2
[
tan
−
1
3
t
2
]
0
∞
=
π
9
×
3
2
×
π
2
=
π
2
12
© examsnet.com
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