Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^(pi/4) sec^2 x "d"x`
Advertisements
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^(π/4) cot^2x.dx`
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate: `int_0^(pi/2) x sin 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_(pi/5)^((3pi)/10) sinx/(sinx + cosx) "d"x` =
`int_0^1 (x^2 - 2)/(x^2 + 1) "d"x` =
Let I1 = `int_"e"^("e"^2) 1/logx "d"x` and I2 = `int_1^2 ("e"^x)/x "d"x` then
Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x "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^(pi/2) (sin2x)/(1 + sin^2x) "d"x`
Evaluate: `int_0^1(x + 1)^2 "d"x`
Evaluate: `int_(pi/6)^(pi/3) sin^2 x "d"x`
Evaluate: `int_0^(pi/4) (tan^3x)/(1 + cos 2x) "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_0^1 1/sqrt(3 + 2x - x^2) "d"x`
Evaluate: `int_0^1 x* tan^-1x "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^(1/2) 1/((1 - 2x^2) sqrt(1 - x^2)) "d"x`
Evaluate: `int_0^pi x*sinx*cos^2x* "d"x`
Evaluate: `int_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
Evaluate:
`int_(π/4)^(π/2) cot^2x dx`.
Evaluate: `int_0^1 tan^-1(x/sqrt(1 - x^2))dx`.
Evaluate:
`int_0^(π/2) sin^8x dx`
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
The value of `int_2^(π/2) sin^3x dx` = ______.
`int_0^1 x^2/(1 + x^2)dx` = ______.
Evaluate `int_(-π/2)^(π/2) sinx/(1 + cos^2x)dx`
If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.
