Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{Let, }I = \int_0^\frac{\pi}{2} \log \tan x\ dx ...................(1)\]
\[ = \int_0^\frac{\pi}{2} \log \tan\left( \frac{\pi}{2} - x \right) dx ....................\left[ Using, \int_0^a f\left( x \right) dx = \int_0^a f\left( a - x \right) dx \right]\]
\[ = \int_0^\frac{\pi}{2} \log cot x\ dx ....................(2)\]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_0^\frac{\pi}{2} \log \tan x d x + \int_0^\frac{\pi}{2} \log cotx\ dx\]
\[ = \int_0^\frac{\pi}{2} \log\left( \tan x \times cotx \right)dx\]
\[ = \int_0^\frac{\pi}{2} \log1 dx = 0\]
\[\text{Hence, }I = 0\]
APPEARS IN
संबंधित प्रश्न
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
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
What is the meaning of a definite integral?
