English

Choose the correct option from the given alternatives : ∫0π2sin2x⋅dx(1+cosx)2 = ______. - Mathematics and Statistics

Advertisements
Advertisements

Question

Choose the correct option from the given alternatives : 

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

Options

  • `(4 - pi)/2`

  • `(pi - 4)/2`

  • `4 - pi/(2)`

  • `(4 + pi)/2`

MCQ
Fill in the Blanks
Advertisements

Solution

`int_0^(pi/2) (sin^2x*dx)/(1 + cosx)^2` = `bb(underline((4 - pi)/2))`.

Explanation:

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

=  `int_0^(pi/2) ((1 + cosx)(1-cosx))/(1 + cosx)^2dx` 

=  `int_0^(pi/2) (1-cosx)/(1+cosx)dx` 

= `int_0^(pi/2) (2sin^2  x/2)/(2cos^2  x/2)dx`

= `int_0^(pi/2)tan^2  x/2dx`

= `int_0^(pi/2) (sec^2  x/2-1)dx`

= `(tan  x/2)/(1/2) - x`

= `2[tan  x/2-x]_0^(pi/2)`

= `2[tan  pi/4-pi/2]`

= `2 - pi/2`

= `(4-pi)/2`

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
  Is there an error in this question or solution?
Chapter 4: Definite Integration - Miscellaneous Exercise 4 [Page 175]

RELATED QUESTIONS

Evaluate: `int_0^(π/4) cot^2x.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`


Let I1 = `int_"e"^("e"^2)  1/logx  "d"x` and I2 = `int_1^2 ("e"^x)/x  "d"x` then 


`int_0^(pi/2) log(tanx)  "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/4)  cosx/(4 - sin^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_(-4)^2 1/(x^2 + 4x + 13)  "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_0^3 x^2 (3 - x)^(5/2)  "d"x`


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


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_0^1 (1/(1 + x^2)) sin^-1 ((2x)/(1 + x^2))  "d"x`


Evaluate: `int_0^(pi/4)  (cos2x)/(1 + cos 2x + sin 2x)  "d"x`


Evaluate: `int_0^(π/4) sec^4 x  dx`


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


Evaluate:

`int_0^(π/2) sin^8x  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) (sin 2x)/(1 + sin^4x)dx`


`int_0^1 x^2/(1 + x^2)dx` = ______.


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×