मराठी

D∫tan-1x dx is equal to ______.

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • `(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
रिकाम्या जागा भरा
Advertisements

उत्तर

`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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise [पृष्ठ १६७]

APPEARS IN

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Prove that:

`int sqrt(x^2 - a^2)dx = x/2sqrt(x^2 - a^2) - a^2/2log|x + sqrt(x^2 - a^2)| + c`


Integrate the function in xlog x.


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


Integrate the function in `e^x (1 + sin x)/(1+cos x)`.


Prove that:

`int sqrt(x^2 + a^2)dx = x/2 sqrt(x^2 + a^2) + a^2/2 log |x + sqrt(x^2 + a^2)| + c`


Evaluate the following : `int x^2tan^-1x.dx`


Evaluate the following : `int x^2*cos^-1 x*dx`


Evaluate the following : `int cos(root(3)(x)).dx`


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

sin (log x)


Integrate the following functions w.r.t. x : `e^x .(1/x - 1/x^2)`


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

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


Integrate the following functions w.r.t. x : `log(1 + x)^((1 + x)`


If f(x) = `sin^-1x/sqrt(1 - x^2), "g"(x) = e^(sin^-1x)`, then `int f(x)*"g"(x)*dx` = ______.


Choose the correct options from the given alternatives :

`int (1)/(cosx - cos^2x)*dx` =


Choose the correct options from the given alternatives :

`int [sin (log x) + cos (log x)]*dx` =


Integrate the following w.r.t. x: `(1 + log x)^2/x`


Integrate the following w.r.t.x : `log (1 + cosx) - xtan(x/2)`


Evaluate the following.

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


Evaluate the following.

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


Evaluate: `int "dx"/("9x"^2 - 25)`


Evaluate: `int e^x/sqrt(e^(2x) + 4e^x + 13)` dx


Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`


`int 1/sqrt(2x^2 - 5)  "d"x`


Choose the correct alternative:

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


`int ("d"x)/(x - x^2)` = ______


`int [(log x - 1)/(1 + (log x)^2)]^2`dx = ?


Evaluate the following:

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


Evaluate the following:

`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`


Evaluate the following:

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


The value of `int_0^(pi/2) log ((4 + 3 sin x)/(4 + 3 cos x))  dx` is


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


`int(logx)^2dx` equals ______.


`int_0^1 x tan^-1 x  dx` = ______.


Evaluate: 

`int(1+logx)/(x(3+logx)(2+3logx))  dx`


`int(xe^x)/((1+x)^2)  dx` = ______


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


Evaluate:

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


Evaluate the following.

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


`∫ sin^(−1)` xdx is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×