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Byju's Answer
Standard XII
Mathematics
Standard Formulae - 1
Evaluate the ...
Question
Evaluate the following integrals:
∫
π
6
π
3
1
1
+
cot
3
2
x
d
x
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Solution
Let I =
∫
π
6
π
3
1
1
+
cot
3
2
x
d
x
.....(1)
Then,
I
=
∫
π
6
π
3
1
1
+
cot
3
2
π
3
+
π
6
-
x
d
x
∫
a
b
f
x
d
x
=
∫
a
b
f
a
+
b
-
x
d
x
=
∫
π
6
π
3
1
1
+
cot
3
2
π
2
-
x
d
x
=
∫
π
6
π
3
1
1
+
tan
3
2
x
d
x
=
∫
π
6
π
3
cot
3
2
x
cot
3
2
x
+
1
d
x
.
.
.
.
.
2
Adding (1) and (2), we get
2
I
=
∫
π
6
π
3
1
+
cot
3
2
x
1
+
cot
3
2
x
d
x
⇒
2
I
=
∫
π
6
π
3
d
x
⇒
2
I
=
x
π
6
π
3
⇒
2
I
=
π
3
-
π
6
=
π
6
⇒
I
=
π
12
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