Advertisements
Advertisements
प्रश्न
Evaluate: `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2) "d"x`
उत्तर
Let I = `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2) "d"x`
Put sin−1x = t
∴ x = sin t
∴ dx = cos t dt
When x = 0, t = 0 and when x = `1/sqrt(2)`, t = `pi/4`
∴ I = `int_0^(pi/4) "t"/(1 - sin^2"t")^(3/2) xx cos "t" "dt"`
= `int_0^(pi/4) "t"/(cot^2 "t")^(3/2) xx cos "t" "dt"`
= `int_0^(pi/4) "t" sec^2 "t" "dt"`
= `["t" int sec^2 "t" "dt"]_0^(pi/4) - int_0^(pi/4)["d"/("dt") ("t") int sec^2 "t" "dt"]"dt"`
= `["t"*tan "t"]_0^(pi/4) - int_0^(pi/4)1* tan "t" "dt"`
= `(pi/4* tan pi/4 - 0) - [log|sec "t"|]_0^(pi/4)`
= `pi/4(1) - [log|sec pi/4| -log|sec 0|]`
= `p/4 - (log sqrt(2) - log 1)`
= `pi/4 - (log 2^(1/2) - 0)`
∴ I = `pi/4 - 1/2 log 2`
APPEARS IN
संबंधित प्रश्न
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate: `int_0^(pi/2) x sin x.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_(pi/5)^((3pi)/10) sinx/(sinx + cosx) "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^(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^(pi/2) (sin2x)/(1 + sin^2x) "d"x`
Evaluate: `int_(pi/6)^(pi/3) sin^2 x "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_0^(pi/4) cosx/(4 - sin^2 x) "d"x`
Evaluate: `int_0^(pi/2) (sin^2x)/(1 + cos x)^2 "d"x`
Evaluate: `int_0^(pi/2) (sin^4x)/(sin^4x + cos^4x) "d"x`
Evaluate: `int_(-1)^1 |5x - 3| "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_0^1 x* tan^-1x "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_(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`
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_0^1 tan^-1(x/sqrt(1 - x^2))dx`.
The value of `int_2^(π/2) sin^3x dx` = ______.
Evaluate:
`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`
Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`
