Advertisements
Advertisements
प्रश्न
Evaluate the definite integral:
`int_0^1 dx/(1+x^2)`
उत्तर
`int_0^1 dx/(1 + x^2)`
`= [tan^-1 x]_0^1`
`= tan^-1 (1) - tan^-1 0`
`= pi/4 - 0`
`= pi/4`
APPEARS IN
संबंधित प्रश्न
Evaluate : `∫_0^(π/2)(sin^2 x)/(sinx+cosx)dx`
Evaluate :`int_0^(pi/2)(2^(sinx))/(2^(sinx)+2^(cosx))dx`
Evaluate the definite integral:
`int_2^3 1/x dx`
Evaluate the definite integral:
`int_1^2 (4x^3 - 5x^2 + 6x + 9) dx`
Evaluate the definite integral:
`int_0^(pi/4) sin2xdx`
Evaluate the definite integral:
`int_0^(pi/2) cos 2x dx`
Evaluate the definite integral:
`int_4^5 e^x dx`
Evaluate the definite integral:
`int_0^(pi/4) tan x dx`
Evaluate the definite integral:
`int_(pi/6)^(pi/4) cosec x dx`
Evaluate the definite integral:
`int_0^1 dx/sqrt(1-x^2)`
Evaluate the definite integral:
`int_2^3 dx/(x^2 - 1)`
Evaluate the definite integral:
`int_0^(pi/2) cos^2 xdx`
Evaluate the definite integral:
`int_0^1 (2x + 3)/(5x^2 + 1) dx`
Evaluate the definite integral:
`int_0^1 x e^(x^2) dx`
Evaluate the definite integral:
`int_1^2 (5x^2)/(x^2 + 4x + 3)`
Evaluate the definite integral:
`int_0^(pi/4) (2 sec^2 x + x^3 + 2) dx`
Evaluate the definite integral:
`int_0^pi (sin^2 x/2 - cos^2 x/2) dx`
Evaluate the definite integral:
`int_0^2 (6x +3)/(x^2 + 4)` dx
Evaluate the definite integral:
`int_0^1 (xe^x + sin (pix)/4)`
`int_6^(2/3) (dx)/(4 + 9x^2)` equals
Evaluate:
`int_0^π(sin^4x + cos^4x)dx`
Hence evaluate:
`int_(-2π)^(2π) (sin^4x + cos^4x)/(1 + e^x)dx`
