मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Evaluate: ∫0π2cosx(1+sinx)(2+sinx) dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

Let I = `int_0^(pi/2) cos x/((1 + sinx)(2 + sinx))  "d"x`

Put sin x = t

∴ cos x dx = dt

When x = 0, t = 0 and when x = `pi/2`, t = 1

∴ I = `int_0^1 "dt"/((1 + "t")(2 + "t"))`

Let `1/((1 + "t")(2 + "t")) = "A"/(1 + "t") + "B"/(2 + "t")` ........(i)

∴ 1 = A(2 + t) + B(1 + t)  ........(ii)

Putting t = −1 in (ii), we get

A = 1

Putting t = −2 in (ii), we get

1 = − B

∴ B = −1

 From (i), we get

`1/((1 + "t")(2 + "t")) = 1/(1 + "t") - 1/(2 + "t")`

∴ I = `int_0^1(1/(1 + "t") - 1/(2 + "t")) "dt"`

= `int_0^1 1/(1 + "t")  "dt" - int_0^1 1/(2 + 1)  "dt"`

= `[log|1 + "t"|]_0^1 - [log|2 + "t"|]_0^1`

= (log 2 − log 1) − (log 3 − log 2)

= `log 2 - 0 - log(3/2)`

= `log(2 xx 2/3)`

∴ I = `log(4/3)`

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2.4: Definite Integration - Short Answers II

संबंधित प्रश्‍न

Evaluate: `int_0^(π/4) cot^2x.dx`


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`


Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.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` = ______.


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` =


`int_0^1 (x^2 - 2)/(x^2 + 1)  "d"x` =


`int_0^(pi/2) log(tanx)  "d"x` =


Evaluate: `int_0^1 1/(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_0^1(x + 1)^2  "d"x`


Evaluate:

`int_0^(pi/2) cos^3x  dx`


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_1^3 (cos(logx))/x  "d"x`


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


Evaluate: `int_3^8 (11 - x)^2/(x^2 + (11 - x)^2)  "d"x`


Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13)  "d"x`


Evaluate: `int_0^(pi/2) 1/(5 + 4cos x)  "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^(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_(-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/4) log(1 + tanx)  "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` = ______.


If `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`, then ______.


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


Evaluate:

`int_0^(π/2) sin^8x  dx`


Evaluate:

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


Evaluate `int_(-π/2)^(π/2) sinx/(1 + cos^2x)dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×