English

D∫tan-1x dx is equal to ______. - Mathematics

Advertisements
Advertisements

Question

`int tan^-1 sqrt(x)  "d"x` is equal to ______.

Options

  • `(x + 1) tan^-1 sqrt(x) - sqrt(x) + "C"`

  • `x tan^-1 sqrt(x) - sqrt(x) + "C"`

  • `sqrt(x) - x tan^-1 sqrt(x) + "C"`

  • `sqrt(x) - (x + 1) tan^-1 sqrt(x) + "C"`

MCQ
Fill in the Blanks
Advertisements

Solution

`int tan^-1 sqrt(x)  "d"x` is equal to `(x + 1) tan^-1 sqrt(x) - sqrt(x) + "C"`.

Explanation:

Let I = `int 1 * tan^-1 sqrt(x)  "d"x`

= `tan^-1 sqrt(x) int 1 "d"x - int[(tan^-1  sqrt(x))"'" int 1"d"x]"d"x`

= `tan^-1 sqrt(x) * x - int 1/(1 + x) * 1/(2sqrt(x)) * x"d"x`  ....[Integrating by parrts]

= `xtan^-1 sqrt(x) - 1/2 int sqrt(x)/(1 + x) "d"x`

Put x = t2

⇒ dx = 2t dt

∴ I = `xtan^-1 sqrt(x) - int "t"^2/(1 + "t"^2) "d"x`

= `xtan^-1 sqrt(x) - int (1 - 1/(1 + "t"^2))"dt"`

= `xtan^-1 sqrt(x) - "t" + tan^-1 1 + "C"`

= `xtan^-1 sqrt(x) - sqrt(x) + tan^-1 sqrt(x) + "C"`

= `(x + 1) tan^-1 sqrt(x) - sqrt(x) + "C"`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise [Page 167]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 7 Integrals
Exercise | Q 50 | Page 167

RELATED QUESTIONS

Integrate : sec3 x w. r. t. x.


If u and v are two functions of x then prove that

`intuvdx=uintvdx-int[du/dxintvdx]dx`

Hence evaluate, `int xe^xdx`


Integrate the function in x cos-1 x.


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


Integrate the function in `(x cos^(-1) x)/sqrt(1-x^2)`.


Integrate the function in `((x- 3)e^x)/(x - 1)^3`.


Find : 

`∫(log x)^2 dx`


Choose the correct options from the given alternatives :

`int (1)/(x + x^5)*dx` = f(x) + c, then `int x^4/(x + x^5)*dx` =


Choose the correct options from the given alternatives :

`int cos -(3)/(7)x*sin -(11)/(7)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 : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`


Choose the correct alternative from the following.

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


Evaluate: `int "dx"/(3 - 2"x" - "x"^2)`


`int ["cosec"(logx)][1 - cot(logx)]  "d"x`


`int (cos2x)/(sin^2x cos^2x)  "d"x`


`int sin4x cos3x  "d"x`


`int"e"^(4x - 3) "d"x` = ______ + c


Evaluate `int (2x + 1)/((x + 1)(x - 2))  "d"x`


`int "e"^x x/(x + 1)^2  "d"x`


The value of `int "e"^(5x) (1/x - 1/(5x^2))  "d"x` is ______.


`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`


Find: `int e^x.sin2xdx`


Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.


Solve: `int sqrt(4x^2 + 5)dx`


If `int(x + (cos^-1 3x)^2)/sqrt(1 - 9x^2)dx = 1/α(sqrt(1 - 9x^2) + (cos^-1 3x)^β) + C`, where C is constant of integration , then (α + 3β) is equal to ______.


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


`int((4e^x - 25)/(2e^x - 5))dx = Ax + B log(2e^x - 5) + c`, then ______.


Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.


`int1/(x+sqrt(x))  dx` = ______


`int logx  dx = x(1+logx)+c`


Evaluate `int(1 + x + (x^2)/(2!))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^3/sqrt(1+x^4)dx`


Evaluate the following.

`intx^3/sqrt(1+x^4)  dx`


Evaluate the following.

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


Evaluate:

`inte^x "cosec"  x(1 - cot x)dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×