Advertisements
Advertisements
प्रश्न
Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
Advertisements
उत्तर
Let I = `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
∴ I = `int_(-1)^1 1/(9 - x^2) "d"x + int_(-1)^1 x^3/(9 - x^2) "d"x`
= I1 + I2 ...(say) ......(i)
Let `"f"(x) = 1/(9 - x^2)`
∴ f(−x) = `1/(9 - (- x)^2`
= `1/(9 - x^2)`
= f(x)
∴ f(x) is an even function.
∴ I1 = `int_(-1)^1 1/(9 - x^2) "d"x`
= `2 int_0^1 1/(9 - x^2) "d"x`
= `2 int_0^1 1/(3^2 - x^2) "d"x`
= `2[1/(2 xx 3)* log|(3 + x)/(3 - x)|]_0^1`
= `1/3[log(4/2) - log(1)]`
∴ I1 = `1/3 log 2`
Let g(x) = `x^3/(9 - x^2)`
∴ g(−x) = `(-x)^3/(9 - (- x)^2`
= `(-x^3)/(9 - x^2)`
= − g(x)
∴ g (x) is an odd function.
∴ I2 = `int_(-1)^1 x^3/(9 - x^2) "d"x` = 0
From (i), we get
I = I1 + I2
∴ I = `1/3 log 2 + 0`
∴ I = `1/3 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`
Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`
`int_0^(x/4) sqrt(1 + sin 2x) "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` =
Evaluate: `int_0^1 |x| "d"x`
Evaluate: `int_0^1 1/sqrt(1 - x^2) "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/4) (tan^3x)/(1 + cos 2x) "d"x`
Evaluate: `int_0^(pi/4) cosx/(4 - sin^2 x) "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_(-1)^1 |5x - 3| "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_0^(pi/4) sec^4x "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_(1/sqrt(2))^1 (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2) "d"x`
Evaluate: `int_0^1 (log(x + 1))/(x^2 + 1) "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^pi 1/(3 + 2sinx + cosx) "d"x`
`int_0^(π/2) sin^6x cos^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` = ______.
Find the value of ‘a’ if `int_2^a (x + 1)dx = 7/2`
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)`
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.
