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^oo xe^-x.dx`
Choose the correct option from the given alternatives :
`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = ______.
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` =
Evaluate: `int_(pi/6)^(pi/3) cosx "d"x`
Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x "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_(pi/6)^(pi/3) sin^2 x "d"x`
Evaluate: `int_0^pi cos^2 x "d"x`
Evaluate: `int_1^3 (cos(logx))/x "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_(-1)^1 |5x - 3| "d"x`
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "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_0^3 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/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_0^1 (log(x + 1))/(x^2 + 1) "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_0^(pi/4) log(1 + tanx) "d"x`
`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_0^1 tan^-1(x/sqrt(1 - x^2))dx`.
Evaluate:
`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`
Find the value of ‘a’ if `int_2^a (x + 1)dx = 7/2`
Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`
