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
Prove that:
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_2^3 e^{- x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
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_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
Γ(1) is
Choose the correct alternative:
`Γ(3/2)`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
