Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^1 1/(1 + x^2) "d"x`
Advertisements
उत्तर
`int_0^1 1/(1 + x^2) "d"x = [tan^-1 x]_0^1`
= tan–11 – tan–10
= `pi/4 - 0`
= `pi/4`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int_0^(π/4) cot^2x.dx`
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate: `int_0^oo xe^-x.dx`
Evaluate the following:
`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`
Choose the correct option from the given alternatives :
`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = ______.
Let I1 = `int_"e"^("e"^2) 1/logx "d"x` and I2 = `int_1^2 ("e"^x)/x "d"x` then
`int_0^4 1/sqrt(4x - x^2) "d"x` =
`int_0^(pi/2) log(tanx) "d"x` =
Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x "d"x`
Evaluate: `int_0^1 |x| "d"x`
Evaluate: `int_0^1 1/sqrt(1 - x^2) "d"x`
Evaluate: `int_1^2 x/(1 + x^2) "d"x`
Evaluate: `int_(pi/6)^(pi/3) sin^2 x "d"x`
Evaluate: `int_0^(pi/2) sqrt(1 - cos 4x) "d"x`
Evaluate:
`int_0^(pi/2) cos^3x dx`
Evaluate: `int_1^3 (cos(logx))/x "d"x`
Evaluate: `int_0^(pi/2) (sin^2x)/(1 + cos x)^2 "d"x`
Evaluate: `int_0^9 sqrt(x)/(sqrt(x) + sqrt(9 - x) "d"x`
Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x`
Evaluate: `int_(-1)^1 |5x - 3| "d"x`
Evaluate: `int_0^1 1/sqrt(3 + 2x - x^2) "d"x`
Evaluate: `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2) "d"x`
Evaluate: `int_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-x)) "d"x`
Evaluate: `int_0^3 x^2 (3 - x)^(5/2) "d"x`
Evaluate: `int_0^(pi/4) (sec^2x)/(3tan^2x + 4tan x + 1) "d"x`
Evaluate: `int_0^1 (log(x + 1))/(x^2 + 1) "d"x`
Evaluate: `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
Evaluate: `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
Evaluate: `int_0^1 tan^-1(x/sqrt(1 - x^2))dx`.
Evaluate:
`int_0^(π/2) sin^8x dx`
Evaluate:
`int_-4^5 |x + 3|dx`
The value of `int_2^(π/2) sin^3x dx` = ______.
Evaluate:
`int_(π/6)^(π/3) (root(3)(sinx))/(root(3)(sinx) + root(3)(cosx))dx`
Evaluate:
`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`
`int_0^1 x^2/(1 + x^2)dx` = ______.
If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.
