Advertisements
Advertisements
प्रश्न
Evaluate: `int_-2^1 sqrt(5 - 4x - x^2)dx`
Advertisements
उत्तर
Let I = `int_-2^1 sqrt(5 - 4x - x^2)dx`
= `int_-2^1 sqrt(-(x^2 + 4x - 5))dx`
= `int_2^1 sqrt(-(x^2 + 4x + 2^2 - 2^2 - 5))dx`
= `int_-2^1 sqrt(-{(x + 2)^2 - 9})dx`
= `int_-2^1 sqrt(3^2 - (x + 2)^2)dx`
= `[(x + 2)/2 sqrt(3^2 - (x + 2)^2) + 3^2/2 sin^-1 ((x + 2)/3)]_-2^1`
= `0 + 9/2 . π/2 - (0 + 0)`
= `(9π)/4`
संबंधित प्रश्न
Evaluate : `int x^2/((x^2+2)(2x^2+1))dx`
Integrate the rational function:
`1/(x^4 - 1)`
Integrate the rational function:
`1/(e^x -1)`[Hint: Put ex = t]
`int (xdx)/((x - 1)(x - 2))` equals:
Find :
`∫ sin(x-a)/sin(x+a)dx`
Integrate the following w.r.t. x : `(12x^2 - 2x - 9)/((4x^2 - 1)(x + 3)`
Integrate the following w.r.t. x : `(1)/(x(x^5 + 1)`
Integrate the following w.r.t. x : `(1)/(x^3 - 1)`
Integrate the following with respect to the respective variable : `(6x + 5)^(3/2)`
Integrate the following w.r.t.x : `(1)/((1 - cos4x)(3 - cot2x)`
Integrate the following w.r.t.x : `sqrt(tanx)/(sinx*cosx)`
Evaluate: `int (2"x" + 1)/(("x + 1")("x - 2"))` dx
For `int ("x - 1")/("x + 1")^3 "e"^"x" "dx" = "e"^"x"` f(x) + c, f(x) = (x + 1)2.
`int "e"^(3logx) (x^4 + 1)^(-1) "d"x`
`int 1/(x(x^3 - 1)) "d"x`
If f'(x) = `x - 3/x^3`, f(1) = `11/2` find f(x)
`int 1/(4x^2 - 20x + 17) "d"x`
`int (sinx)/(sin3x) "d"x`
`int "e"^x ((1 + x^2))/(1 + x)^2 "d"x`
`int x^3tan^(-1)x "d"x`
`int xcos^3x "d"x`
`int ((2logx + 3))/(x(3logx + 2)[(logx)^2 + 1]) "d"x`
`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5) "dt"`
If `intsqrt((x - 7)/(x - 9)) dx = Asqrt(x^2 - 16x + 63) + log|x - 8 + sqrt(x^2 - 16x + 63)| + c`, then A = ______
Verify the following using the concept of integration as an antiderivative
`int (x^3"d"x)/(x + 1) = x - x^2/2 + x^3/3 - log|x + 1| + "C"`
`int 1/(x^2 + 1)^2 dx` = ______.
If `intsqrt((x - 5)/(x - 7))dx = Asqrt(x^2 - 12x + 35) + log|x| - 6 + sqrt(x^2 - 12x + 35) + C|`, then A = ______.
Evaluate: `int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx`
Evaluate:
`int (x + 7)/(x^2 + 4x + 7)dx`
