हिंदी

Evaluate: ∫-111a2ex+b2e-x - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate: `int_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-x))  "d"x`

योग
Advertisements

उत्तर

Let I = `int_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-x))  "d"x`

= `int_(-1)^1 1/("a"^2"e"^x + ("b"^2)/("e"^x))  "d"x`

= `int_(-1)^1 "e"^x/("a"^2("e"^x)^2 + "b"^2)  "d"x`

Put ex = t

∴ ex dx = dt

When x = −1, t = e−1 and when x = 1, t = e

∴ I = `int_("e"^-1)^"e"  "dt"/("a"^2"t"^2 + "b"^2)`

= `1/("a"^2) int_("e"^-1)^"e"  "dt"/("t"^2 + ("b"/"a")^2`

= `1/("a"^2)[1/("b"/"a")tan^-1 ("t"/("b"/"a"))]_("e"^-1)^"e"`

= `1/"ab"[tan^-1("at"/"b")]_("e"^-1)^"e"`

∴ I = `1/"ab"[tan^-1("ae"/"b") - tan^-1("a"/"be")]`

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2.4: Definite Integration - Short Answers II

संबंधित प्रश्न

Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`


Evaluate: `int_0^(pi/2) x sin x.dx`


Evaluate the following:

`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`


Choose the correct option from the given alternatives : 

`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = ______.


`int_0^1 (x^2 - 2)/(x^2 + 1)  "d"x` =


`int_0^4 1/sqrt(4x - x^2)  "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 1/sqrt(1 - x^2)  "d"x`


Evaluate: `int_0^1 "e"^x/sqrt("e"^x - 1)  "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/2) cos^3x  dx`


Evaluate: `int_0^pi cos^2 x  "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^(pi/2) cos x/((1 + sinx)(2 + sinx))  "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^pi x*sinx*cos^2x* "d"x`


Evaluate: `int_(-1)^1 (1 + x^2)/(9 - 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` = ______.


Evaluate `int_(π/6)^(π/3) cos^2x  dx`


Evaluate:

`int_-4^5 |x + 3|dx`


The value of `int_2^(π/2) sin^3x  dx` = ______.


Evaluate:

`int_(π/6)^(π/3) (root(3)(sinx))/(root(3)(sinx) + root(3)(cosx))dx`


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)`


If `int_0^π f(sinx)dx = kint_0^π f(sinx)dx`, then find the value of k.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×