Advertisements
Advertisements
प्रश्न
`int_0^1 (x^2 - 2)/(x^2 + 1) "d"x` =
विकल्प
`1 - (3pi)/4`
`2 - (3pi)/4`
`1 + (3pi)/4`
`2 + (3pi)/4`
उत्तर
`1 - (3pi)/4`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int_0^oo xe^-x.dx`
Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`
Evaluate the following:
`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`
If `int_0^1 ("d"x)/(sqrt(1 + x) - sqrt(x)) = "k"/3`, then k is equal to ______.
`int_0^4 1/sqrt(4x - x^2) "d"x` =
Evaluate: `int_0^(pi/4) sec^2 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_0^1 "e"^x/sqrt("e"^x - 1) "d"x`
Evaluate: `int_0^1(x + 1)^2 "d"x`
Evaluate: `int_0^(pi/2) cos^3x "d"x`
Evaluate: `int_0^pi cos^2 x "d"x`
Evaluate: `int_0^(pi/4) cosx/(4 - sin^2 x) "d"x`
Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x`
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "d"x`
Evaluate: `int_0^(pi/4) sec^4x "d"x`
Evaluate: `int_0^(pi/2) 1/(5 + 4cos x) "d"x`
Evaluate: `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx)) "d"x`
Evaluate: `int_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-x)) "d"x`
Evaluate: `int_0^"a" 1/(x + sqrt("a"^2 - x^2)) "d"x`
Evaluate: `int_0^3 x^2 (3 - x)^(5/2) "d"x`
Evaluate: `int_(1/sqrt(2))^1 (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2) "d"x`
Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
Evaluate: `int_0^1 (1/(1 + x^2)) sin^-1 ((2x)/(1 + x^2)) "d"x`
Evaluate: `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
Evaluate: `int_0^(pi/4) log(1 + tanx) "d"x`
Evaluate:
`int_(π/4)^(π/2) cot^2x dx`.
Evaluate:
`int_0^(π/2) sin^8x dx`
Evaluate `int_(π/6)^(π/3) cos^2x dx`
The value of `int_2^(π/2) sin^3x dx` = ______.
Evaluate:
`int_0^(π/2) sinx/(1 + cosx)^3 dx`
Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`
If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.