Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^(pi/4) sec^2 x "d"x`
उत्तर
`int_0^(pi/4) sec^2 x "d"x = [tan x]_0^(pi/4)`
= `tan pi/4 - tan 0`
= 1 − 0
= 1
APPEARS IN
संबंधित प्रश्न
Evaluate: `int_0^oo xe^-x.dx`
Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`
Choose the correct option from the given alternatives :
`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = ______.
`int_(pi/5)^((3pi)/10) sinx/(sinx + cosx) "d"x` =
`int_0^1 (x^2 - 2)/(x^2 + 1) "d"x` =
`int_0^4 1/sqrt(4x - x^2) "d"x` =
Evaluate: `int_(pi/6)^(pi/3) cosx "d"x`
Evaluate: `int_0^1 |x| "d"x`
Evaluate: `int_0^1 "e"^x/sqrt("e"^x - 1) "d"x`
Evaluate: `int_0^(pi/2) sqrt(1 - cos 4x) "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) (tan^3x)/(1 + cos 2x) "d"x`
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_(-1)^1 |5x - 3| "d"x`
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "d"x`
Evaluate: `int_0^(pi/2) 1/(5 + 4cos 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_0^1 "t"^2 sqrt(1 - "t") "dt"`
Evaluate: `int_0^(1/2) 1/((1 - 2x^2) sqrt(1 - x^2)) "d"x`
Evaluate: `int_(1/sqrt(2))^1 (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2) "d"x`
Evaluate: `int_0^pi x*sinx*cos^2x* "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_0^1 tan^-1(x/sqrt(1 - x^2))dx`.
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
Evaluate:
`int_-4^5 |x + 3|dx`
Evaluate:
`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`
`int_0^1 x^2/(1 + x^2)dx` = ______.
Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`
Evaluate `int_(-π/2)^(π/2) sinx/(1 + cos^2x)dx`
