Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{Let, }I = \int_0^\frac{\pi}{2} \log \tan x\ dx ...................(1)\]
\[ = \int_0^\frac{\pi}{2} \log \tan\left( \frac{\pi}{2} - x \right) dx ....................\left[ Using, \int_0^a f\left( x \right) dx = \int_0^a f\left( a - x \right) dx \right]\]
\[ = \int_0^\frac{\pi}{2} \log cot x\ dx ....................(2)\]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_0^\frac{\pi}{2} \log \tan x d x + \int_0^\frac{\pi}{2} \log cotx\ dx\]
\[ = \int_0^\frac{\pi}{2} \log\left( \tan x \times cotx \right)dx\]
\[ = \int_0^\frac{\pi}{2} \log1 dx = 0\]
\[\text{Hence, }I = 0\]
APPEARS IN
RELATED QUESTIONS
Prove that:
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Find: `int logx/(1 + log x)^2 dx`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
