Advertisements
Advertisements
Question
Evaluate : `int_-1^1 log ((2 - x)/(2 + x))dx`.
Advertisements
Solution
Let f(x) = `log((2 - x)/(2 + x))`
We have, f(– x) = `log((2 + x)/(2 - x))`
= `-log((2 - x)/(2 + x))`
= – f(x)
So, f(x) is an odd function.
∴ `int_-1^1 log ((2 - x)/(2 + x))dx` = 0.
APPEARS IN
RELATED QUESTIONS
By using the properties of the definite integral, evaluate the integral:
`int_0^(pi/2) sin^(3/2)x/(sin^(3/2)x + cos^(3/2) x) dx`
By using the properties of the definite integral, evaluate the integral:
`int_0^a sqrtx/(sqrtx + sqrt(a-x)) dx`
Evaluate `int_0^(pi/2) cos^2x/(1+ sinx cosx) dx`
Evaluate : `int _0^(pi/2) "sin"^ 2 "x" "dx"`
Evaluate = `int (tan x)/(sec x + tan x)` . dx
Evaluate: `int_0^pi ("x"sin "x")/(1+ 3cos^2 "x") d"x"`.
Find : `int_ (2"x"+1)/(("x"^2+1)("x"^2+4))d"x"`.
Evaluate the following integral:
`int_0^1 x(1 - x)^5 *dx`
`int_"a"^"b" "f"(x) "d"x` = ______
`int_0^{pi/2} log(tanx)dx` = ______
`int_-9^9 x^3/(4 - x^2)` dx = ______
`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = ______.
f(x) = `{:{(x^3/k; 0 ≤ x ≤ 2), (0; "otherwise"):}` is a p.d.f. of X. The value of k is ______
`int_0^{pi/2} cos^2x dx` = ______
`int_-2^1 dx/(x^2 + 4x + 13)` = ______
`int_{pi/6}^{pi/3} sin^2x dx` = ______
`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = ______
`int_0^1 log(1/x - 1) "dx"` = ______.
Find `int_0^(pi/4) sqrt(1 + sin 2x) "d"x`
`int_0^(2"a") "f"(x) "d"x = 2int_0^"a" "f"(x) "d"x`, if f(2a – x) = ______.
`int_0^(pi/2) (sin^"n" x"d"x)/(sin^"n" x + cos^"n" x)` = ______.
Evaluate the following:
`int_(-pi/4)^(pi/4) log|sinx + cosx|"d"x`
Evaluate:
`int_2^8 (sqrt(10 - "x"))/(sqrt"x" + sqrt(10 - "x")) "dx"`
The value of `int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2)) dx` is
Evaluate: `int_1^3 sqrt(x)/(sqrt(x) + sqrt(4) - x) dx`
Evaluate: `int_(-1)^3 |x^3 - x|dx`
Let `int_0^∞ (t^4dt)/(1 + t^2)^6 = (3π)/(64k)` then k is equal to ______.
`int_0^(π/4) x. sec^2 x dx` = ______.
Evaluate: `int_0^(π/4) log(1 + tanx)dx`.
