Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^(pi/2) sqrt(1 - cos 4x) "d"x`
Advertisements
उत्तर
`int_0^(pi/2) sqrt(1 - cos 4x) "d"x`
= `int_0^(pi/2) sqrt(2sin^2 2x) "d"x` .......`[∵ 1 - cos theta = 2 sin^2 theta/2]`
= `sqrt(2) int_0^(pi/2) sin 2x "d"x`
= `sqrt(2)[(-cos 2x)/2]_0^(pi/2)`
= `sqrt(2)/2 [cos 2 pi/2 - cos 0]`
= `-(sqrt(2))/2 [cos pi - cos 0]`
= `-(sqrt(2))/2 (-1 - 1)`
= `sqrt(2)`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int_0^(π/4) cot^2x.dx`
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_0^(x/4) sqrt(1 + sin 2x) "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
`int_0^4 1/sqrt(4x - x^2) "d"x` =
`int_0^(pi/2) log(tanx) "d"x` =
Evaluate: `int_(pi/6)^(pi/3) cosx "d"x`
Evaluate: `int_0^1 1/(1 + x^2) "d"x`
Evaluate: `int_0^(pi/4) sec^2 x "d"x`
Evaluate: `int_0^1 |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_(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/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_0^1 1/sqrt(3 + 2x - x^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_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-x)) "d"x`
Evaluate: `int_0^1 "t"^2 sqrt(1 - "t") "dt"`
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_0^(pi/4) (cos2x)/(1 + cos 2x + sin 2x) "d"x`
Evaluate: `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
Evaluate: `int_0^(π/4) sec^4 x dx`
`int_0^(π/2) sin^6x cos^2x.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`
Evaluate:
`int_-4^5 |x + 3|dx`
Evaluate:
`int_(π/6)^(π/3) (root(3)(sinx))/(root(3)(sinx) + root(3)(cosx))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)`
