हिंदी

Evaluate: ∫ x tan^-1 x dx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following:

`int x tan^-1 x . dx`

Evaluate:

∫ x tan-1 x dx

मूल्यांकन
Advertisements

उत्तर

Let I = `int x tan^-1 x  .  dx`

= `int (tan^-1 x)x  .  dx`

= `(tan^-1 x) int x  .  dx - int[{d/dx(tan^-1 x) intx.dx}]  .  dx`

= `(tan^-1x) (x^2/2) - int (1/(1 + x^2)) (x^2/2)  .  dx`

= `(x^2 tan^-1)/(2) - (1)/(2) int x^2/(x^2 + 1)  .  dx`

= `x^2/(2) tan^-1x - (1)/(2) ((x^2 + 1)-1)/(x^2 + 1)  ⋅  dx`

= `x^2/(2)tan^-1x - (1)/(2)[int(1 - 1/(x^2 + 1))  ⋅  dx]`

= `x^2/(2)tan^-1x - (1)/(2)[int 1  .  dx - int(1)/(x^2 + 1)  .  dx]`

= `x^2/(2)tan^-1 x - (1)/(2)(x - tan^-1x) + c`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Exercise 3.3 [पृष्ठ १३७]

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`


Integrate the function in `x^2e^x`.


Integrate the function in (sin-1x)2.


Integrate the function in tan-1 x.


`intx^2 e^(x^3) dx` equals: 


Evaluate the following : `int x.sin^2x.dx`


Evaluate the following : `int x^3.logx.dx`


Evaluate the following : `int cos sqrt(x).dx`


Evaluate the following : `int sin θ.log (cos θ).dθ`


Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`


Integrate the following functions w.r.t.x:

`e^(5x).[(5x.logx + 1)/x]`


Choose the correct options from the given alternatives :

`int (sin^m x)/(cos^(m+2)x)*dx` = 


Integrate the following with respect to the respective variable : `(sin^6θ + cos^6θ)/(sin^2θ*cos^2θ)`


Integrate the following w.r.t.x : cot–1 (1 – x + x2)


Integrate the following w.r.t.x : log (x2 + 1)


Evaluate the following.

`int x^2 e^4x`dx


Evaluate the following.

`int "e"^"x" "x"/("x + 1")^2` dx


Evaluate the following.

`int "e"^"x" [(log "x")^2 + (2 log "x")/"x"]` dx


Evaluate the following.

`int [1/(log "x") - 1/(log "x")^2]` dx


Choose the correct alternative from the following.

`int (1 - "x")^(-2) "dx"` = 


Evaluate: `int "dx"/(5 - 16"x"^2)`


`int (sinx)/(1 + sin x)  "d"x`


Choose the correct alternative:

`intx^(2)3^(x^3) "d"x` =


`int 1/(x^2 - "a"^2)  "d"x` = ______ + c


`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.


Evaluate the following:

`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`


Evaluate the following:

`int_0^pi x log sin x "d"x`


If u and v ore differentiable functions of x. then prove that:

`int uv  dx = u intv  dx - int [(du)/(d) intv  dx]dx`

Hence evaluate `intlog x  dx`


Find: `int (2x)/((x^2 + 1)(x^2 + 2)) dx`


The integral `int x cos^-1 ((1 - x^2)/(1 + x^2))dx (x > 0)` is equal to ______.


If `π/2` < x < π, then `intxsqrt((1 + cos2x)/2)dx` = ______.


Find `int e^(cot^-1x) ((1 - x + x^2)/(1 + x^2))dx`.


Find: `int e^(x^2) (x^5 + 2x^3)dx`.


Evaluate :

`int(4x - 6)/(x^2 - 3x + 5)^(3/2)  dx`


`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`


Evaluate `int(3x-2)/((x+1)^2(x+3))  dx`


Evaluate:

`int((1 + sinx)/(1 + cosx))e^x dx`


Evaluate:

`inte^x sinx  dx`


Evaluate:

`int (logx)^2 dx`


Evaluate:

`int (sin(x - a))/(sin(x + a))dx`


If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv  dx - int(d/dx u)(intv  dx)dx`. Hence evaluate: `intx cos x  dx`


Complete the following activity:

`int_0^2 dx/(4 + x - x^2) `

= `int_0^2 dx/(-x^2 + square + square)`

= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`

= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`

= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`


Evaluate the following.

`intx^3e^(x^2) dx`


If f′(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).


Evaluate the following.

`int x^3 e^(x^2) dx` 


Evaluate the following.

`intx^2e^(4x)dx`


Evaluate `int(1 + x + x^2/(2!))dx`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×