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AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper
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© examsnet.com
Question : 32
Total: 160
π
∕
3
∫
π
∕
6
1
1
+
√
cotx
d
x
=
π
12
π
6
π
4
π
13
Validate
Solution:
π
/
3
∫
π
/
6
1
1
+
√
cot
x
d
x
=
I
(say)
I
=
π
/
3
∫
π
/
6
√
cos
x
√
sin
x
+
√
cos
x
d
x
...(i)
We have,
b
∫
a
f
(
x
)
d
x
=
b
∫
0
f
(
a
+
b
−
x
)
d
x
a
+
b
−
x
=
π
2
−
x
I
=
π
/
3
∫
π
/
6
√
cos
(
π
2
−
x
)
√
sin
(
π
2
−
x
)
+
√
cos
(
π
2
−
x
)
d
x
I
=
π
/
3
∫
π
/
6
√
sin
x
√
sin
x
+
√
cos
x
d
x
...(ii)
Eqs. (i)
+
(ii)
⇒
2
I
=
π
/
3
∫
π
/
6
(
√
cos
x
√
sin
x
+
√
cos
x
+
√
sin
x
√
sin
x
+
√
cos
x
)
d
x
=
π
/
3
∫
π
/
6
(
√
sin
x
+
√
cos
x
√
sin
x
+
√
cos
x
)
d
x
=
π
/
3
∫
π
/
6
1
⋅
d
x
=
(
x
)
π
/
6
π
/
3
=
(
π
3
−
π
6
)
2
I
=
π
6
⇒
I
=
π
12
© examsnet.com
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