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Definite Integration
NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Definite Integration
NDA 2021 · Mathematics · Definite Integration
What is ∫_0^π ( (x)/(2)) d x equal to?
What is
∫
0
π
ln
(
tan
x
2
)
d
x
\int_0^\pi \ln \left(\tan \frac{x}{2}\right) d x
∫
0
π
ln
(
tan
2
x
)
d
x
equal to?
Options
A
0
B
1
2
\frac{1}{2}
2
1
C
1
D
2
Correct answer
A. 0
Step-by-step solution
I
=
∫
0
π
ln
(
tan
x
2
)
d
x
.
.
.
(
i
)
I
=
∫
0
π
ln
(
tan
(
π
2
−
x
2
)
)
d
x
=
∫
0
π
ln
(
cot
x
2
)
d
x
.
.
.
(
i
i
)
\begin{aligned} \mathrm{I} & =\int_0^\pi \ln \left(\tan \frac{x}{2}\right) d x \qquad \qquad...\mathrm{(i)}\\ \mathrm{I} & =\int_0^\pi \ln \left(\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right) d x \\ & =\int_0^\pi \ln \left(\cot \frac{x}{2}\right) d x \qquad \qquad...\mathrm{(ii)} \end{aligned}
I
I
=
∫
0
π
ln
(
tan
2
x
)
d
x
...
(
i
)
=
∫
0
π
ln
(
tan
(
2
π
−
2
x
)
)
d
x
=
∫
0
π
ln
(
cot
2
x
)
d
x
...
(
ii
)
Adding equation (1) and (2), we get
2
2
2
\mathrm{I}
=
∫
0
π
ln
(
tan
x
2
)
d
x
+
∫
0
π
ln
(
cot
x
2
)
d
x
=\int_0^\pi \ln \left(\tan \frac{x}{2}\right) d x+\int_0^\pi \ln \left(\cot \frac{x}{2}\right) d x
=
∫
0
π
ln
(
tan
2
x
)
d
x
+
∫
0
π
ln
(
cot
2
x
)
d
x
2
I
=
∫
0
π
ln
(
tan
x
2
cot
x
2
)
d
x
2
I
=
∫
0
π
ln
(
tan
x
2
×
1
tan
x
2
)
d
x
I
=
1
2
∫
0
π
ln
(
1
)
d
x
(
∴
ln
1
=
0
)
∴
I
=
0
\begin{aligned} 2 \mathrm{I} & =\int_0^\pi \ln \left(\tan \frac{x}{2} \cot \frac{x}{2}\right) d x \\ 2 \mathrm{I} & =\int_0^\pi \ln \left(\tan \frac{x}{2} \times \frac{1}{\tan \frac{x}{2}}\right) d x \\ \mathrm{I} & =\frac{1}{2} \int_0^\pi \ln (1) d x \quad(\therefore \ln 1=0) \\ & \therefore \mathrm{I}=0\end{aligned}
2
I
2
I
I
=
∫
0
π
ln
(
tan
2
x
cot
2
x
)
d
x
=
∫
0
π
ln
(
tan
2
x
×
tan
2
x
1
)
d
x
=
2
1
∫
0
π
ln
(
1
)
d
x
(
∴
ln
1
=
0
)
∴
I
=
0
$$
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