Advertisements
Advertisements
प्रश्न
Choose the correct option from the given alternatives :
`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = ______.
विकल्प
`(4 - pi)/2`
`(pi - 4)/2`
`4 - pi/(2)`
`(4 + pi)/2`
उत्तर
`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = `bb(underline((4 - pi)/2))`.
Explanation:
`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2 = int_0^(pi/2) (1- cos^2x)/(1 + cosx)^2dx`
= `int_0^(pi/2) ((1 + cosx)(1-cosx))/(1 + cosx)^2dx`
= `int_0^(pi/2) (1-cosx)/(1+cosx)dx`
= `int_0^(pi/2) (2sin^2 x/2)/(2cos^2 x/2)dx`
= `int_0^(pi/2)tan^2 x/2dx`
= `int_0^(pi/2) (sec^2 x/2-1)dx`
= `(tan x/2)/(1/2) - x`
= `2[tan x/2-x]_0^(pi/2)`
= `2[tan pi/4-pi/2]`
= `2 - pi/2`
= `(4-pi)/2`
संबंधित प्रश्न
Evaluate: `int_0^(pi/2) x sin x.dx`
Evaluate: `int_0^oo xe^-x.dx`
Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`
`int_0^(x/4) sqrt(1 + sin 2x) "d"x` =
If `int_0^1 ("d"x)/(sqrt(1 + x) - sqrt(x)) = "k"/3`, then k is equal to ______.
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` =
Evaluate: `int_(pi/6)^(pi/3) cosx "d"x`
Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x "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 "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_3^8 (11 - x)^2/(x^2 + (11 - x)^2) "d"x`
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "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_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^(pi/4) (sec^2x)/(3tan^2x + 4tan x + 1) "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^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
Evaluate: `int_0^(π/4) sec^4 x dx`
If `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`, then ______.
Evaluate:
`int_(π/4)^(π/2) cot^2x dx`.
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
Evaluate `int_(π/6)^(π/3) cos^2x dx`
The value of `int_2^(π/2) sin^3x dx` = ______.
`int_0^1 x^2/(1 + x^2)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.
