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

Evaluate: ∫0π4 cosx4-sin2x dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate: `int_0^(pi/4)  cosx/(4 - sin^2 x)  "d"x`

बेरीज
Advertisements

उत्तर

Let I = `int_0^(pi/4)  cosx/(4 - sin^2 x)  "d"x`

Put sin x = t

∴ cos x dx = dt

When x = 0, t = 0 and when x = `pi/4`,t  `1/sqrt(2)`

∴ I = `int_0^(1/sqrt(2)) "dt"/(4 - "t"^2)`

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

= `[1/(2 xx 2) log|(2 + "t")/(2 - "t")|]_0^(1/sqrt(2))`

= `1/4[log|(2 + 1/sqrt(2))/(2 - 1/sqrt(2))| - log 1]`

= `1/4[log|(2sqrt(2) + 1)/(2sqrt(2) - 1)| - 0]`

∴ I = `1/4 log((2sqrt(2) + 1)/(2sqrt(2) - 1))`

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 I

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`


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^(pi/2) log(tanx)  "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/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_0^9 sqrt(x)/(sqrt(x) + sqrt(9 - x)  "d"x`


Evaluate: `int_3^8 (11 - x)^2/(x^2 + (11 - 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) 1/(5 + 4cos x)  "d"x`


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


Evaluate: `int_0^"a" 1/(x + sqrt("a"^2 - x^2))  "d"x`


Evaluate: `int_0^1 "t"^2 sqrt(1 - "t")  "dt"`


Evaluate: `int_0^(pi/4)  (sec^2x)/(3tan^2x + 4tan x + 1)  "d"x`


Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2)  "d"x`


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


Evaluate: `int_0^(pi/4) log(1 + tanx)  "d"x`


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


Evaluate:

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


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


Evaluate:

`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`


`int_0^1 x^2/(1 + x^2)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`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×