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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Definite Integration
BITSAT 2024 · Mathematics · Definite Integration
The value of definite integral ∫_0^ π / 2 ( x) d x is
The value of definite integral
∫
0
π
/
2
log
(
tan
x
)
d
x
\int_0^{\pi / 2} \log (\tan x) d x
∫
0
π
/2
lo
g
(
tan
x
)
d
x
is
Options
A
0
B
π
4
\frac{\pi}{4}
4
π
C
π
2
\frac{\pi}{2}
2
π
D
π
\pi
π
Correct answer
A. 0
Step-by-step solution
Let
I
=
∫
0
π
2
log
(
tan
x
)
d
x
I=\int_0^{\frac{\pi}{2}} \log (\tan x) d x
I
=
∫
0
2
π
lo
g
(
tan
x
)
d
x
....(i) By applying property,
∫
a
b
f
(
x
)
d
x
=
∫
a
b
f
(
a
+
b
−
x
)
d
x
I
=
∫
0
π
2
log
(
tan
(
π
2
−
x
)
)
d
x
I
=
∫
0
π
2
log
(
cot
x
)
d
x
…
(
i
i
)
\begin{aligned} \int_a^b f(x) d x & =\int_a^b f(a+b-x) d x \\ I & =\int_0^{\frac{\pi}{2}} \log \left(\tan \left(\frac{\pi}{2}-x\right)\right) d x \\ I & =\int_0^{\frac{\pi}{2}} \log (\cot x) d x\ldots(ii) \end{aligned}
∫
a
b
f
(
x
)
d
x
I
I
=
∫
a
b
f
(
a
+
b
−
x
)
d
x
=
∫
0
2
π
lo
g
(
tan
(
2
π
−
x
)
)
d
x
=
∫
0
2
π
lo
g
(
cot
x
)
d
x
…
(
ii
)
Adding Eqs. (i) and (ii),
2
I
=
∫
0
π
2
[
log
(
tan
x
)
+
log
(
cot
x
)
]
d
x
2
I
=
∫
0
π
2
log
1
d
x
=
0
\begin{aligned} & 2 I=\int_0^{\frac{\pi}{2}}[\log (\tan x)+\log (\cot x)] d x \\ & 2 I=\int_0^{\frac{\pi}{2}} \log 1 d x=0 \end{aligned}
2
I
=
∫
0
2
π
[
lo
g
(
tan
x
)
+
lo
g
(
cot
x
)]
d
x
2
I
=
∫
0
2
π
lo
g
1
d
x
=
0
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