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

Evaluate: π∫0π4cot2x.dx

Advertisements
Advertisements

प्रश्न

Evaluate: `int_0^(π/4) cot^2x.dx`

बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Definite Integration - Exercise 4.2 [पृष्ठ १७१]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 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`


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` =


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


Evaluate: `int_(pi/6)^(pi/3) cosx  "d"x`


Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x  "d"x`


Evaluate: `int_0^1 1/(1 + x^2)  "d"x`


Evaluate: `int_1^2 x/(1 + x^2)  "d"x`


Evaluate: `int_0^1 "e"^x/sqrt("e"^x - 1)  "d"x`


Evaluate: `int_0^(pi/2)  (sin2x)/(1 +  sin^2x)  "d"x`


Evaluate: `int_0^1(x + 1)^2  "d"x`


Evaluate: `int_(pi/6)^(pi/3) sin^2 x  "d"x`


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


Evaluate: `int_0^(pi/2) (sin^2x)/(1 + cos x)^2 "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_3^8 (11 - x)^2/(x^2 + (11 - x)^2)  "d"x`


Evaluate: `int_(-1)^1 |5x - 3|  "d"x`


Evaluate: `int_0^(pi/4) sec^4x  "d"x`


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


Evaluate: `int_0^pi 1/(3 + 2sinx + cosx)  "d"x`


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


Evaluate:

`int_0^(π/2) sin^8x  dx`


Evaluate:

`int_(-π/2)^(π/2) |sinx|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`


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


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×