हिंदी

Evaluate: π∫0π4cot2x.dx

Advertisements
Advertisements

प्रश्न

Evaluate: `int_0^(π/4) cot^2x.dx`

योग
Advertisements

उत्तर

`int_0^(π/4) cot^2x.dx`

= `int_0^(π/4) ("cosec"^2x - 1).dx`

= `int_0^(π/4) "cosec"^2x.dx - int_0^(π/4)1.dx`

= `[-cot x]_0^(π/4) - [x]_0^(π/4)`

= `[(-cot π/4) - (-cot 0)] - [π/4 - 0]`

= `-1 + cot 0 - pi/4`

The integral does not exist since cot 0 is not defined.

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Definite Integration - Exercise 4.2 [पृष्ठ १७१]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 4 Definite Integration
Exercise 4.2 | Q 1.03 | पृष्ठ १७१

संबंधित प्रश्न

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


Evaluate: `int_0^oo xe^-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^(x/4) sqrt(1 + sin 2x)  "d"x` =


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` =


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


`int_0^4 1/sqrt(4x - x^2)  "d"x` =


`int_0^(pi/2) log(tanx)  "d"x` =


Evaluate: `int_0^(pi/4) sec^2 x  "d"x`


Evaluate: `int_0^1 |x|  "d"x`


Evaluate: `int_1^2 x/(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) sqrt(1 - cos 4x)  "d"x`


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_(-4)^2 1/(x^2 + 4x + 13)  "d"x`


Evaluate: `int_0^1 1/sqrt(3 + 2x - x^2)  "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_0^3 x^2 (3 - x)^(5/2)  "d"x`


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


`int_0^(π/2) sin^6x cos^2x.dx` = ______.


Evaluate:

`int_0^(π/2) sin^8x  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` = ______.


Find the value of ‘a’ if `int_2^a (x + 1)dx = 7/2`


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)`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×