English

Evaluate: ∫01(11+x2)sin-1(2x1+x2) dx - Mathematics and Statistics

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Let I = `int_0^1 (1/(1 + x^2)) sin^-1 ((2x)/(1 + x^2))  "d"x`

Put x = tan θ

∴ dx = sec2θ dθ

When x = 0, θ = 0 and when x = 1, θ = `pi/4`

∴ I = `int_0^(pi/4)(1/(1 + tan^2 theta)) sin^-1((2tan theta)/(1 + tan^2theta)) sec^2theta  "d"theta`

= `int_0^(pi/4) (1/(sec^2 theta)) sin^-1 (sin 2theta) sec^2theta  "d"theta`

= `int_0^(pi/4) 2theta  "d"theta`

= `2[theta^2/2]_0^(pi/4)`

= `(pi/4)^2 - 0`

∴ I = `pi^2/16`

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
  Is there an error in this question or solution?
Chapter 2.4: Definite Integration - Long Answers III

RELATED QUESTIONS

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^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`


`int_0^(x/4) sqrt(1 + sin 2x)  "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_0^1 1/(1 + x^2)  "d"x`


Evaluate: `int_0^1 "e"^x/sqrt("e"^x - 1)  "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/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^(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_(-4)^2 1/(x^2 + 4x + 13)  "d"x`


Evaluate: `int_0^1 1/sqrt(3 + 2x - x^2)  "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_(-1)^1 1/("a"^2"e"^x + "b"^2"e"^(-x))  "d"x`


Evaluate: `int_(1/sqrt(2))^1  (("e"^(cos^-1x))(sin^-1x))/sqrt(1 - x^2)  "d"x`


Evaluate: `int_0^pi x*sinx*cos^2x* "d"x`


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


Evaluate: `int_0^(π/4) sec^4 x  dx`


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


Evaluate:

`int_(-π/2)^(π/2) |sinx|dx`


Evaluate:

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


Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`


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×