मराठी

Integrate the function in sin-1(2x1+x2).

Advertisements
Advertisements

प्रश्न

Integrate the function in `sin^(-1) ((2x)/(1+x^2))`.

बेरीज
Advertisements

उत्तर

Let `I = sin^-1 ((2x)/ (1 + x^2))  dx`

Put x = tan t

⇒ dx = sec2 t dt

∴ `I = int sin^-1 ((2 tan t)/ (1 + tan^2 t)) sec^2 t dt`

`= int sin^-1 (sin 2t) sec^2 t dt`

`= 2t sec^2 t dt = 2 int sec^2 t dt`

`= 2 {t int sec^2 t dt - int [d/dt(t) * int sec^2 t  dt] dt}`

`= 2 [t tant  - int 1 * tan t  dt]`

= 2 t tan t + 2 log |cos t| + C

`= 2 tan^-1 x*x + 2 log |1/ sqrt (1 + x^2)| + C`     `...[∵ cos t = 1/ (sect) = 1/ (sqrt (1 + tan^2 t)) = 1/ (sqrt (1 + x^2))]`

`= 2 x tan^-1 x + 2 log |(1 + x^2)^(1/2)| + C`

`= 2 x tan^-1 x + 2 (- 1/2) log |1 + x^2| + C`

`= 2 x tan^-1 x - log |1 + x^2| + C`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.6 [पृष्ठ ३२८]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.6 | Q 22 | पृष्ठ ३२८

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 `x^2e^x`.


Integrate the function in x log 2x.


Integrate the function in e2x sin x.


Integrate the following functions w.r.t. x : `xsqrt(5 - 4x - x^2)`


Integrate the following functions w.r.t. x : `sqrt(2x^2 + 3x + 4)`


Integrate the following functions w.r.t. x : cosec (log x)[1 – cot (log x)] 


Choose the correct options from the given alternatives :

`int sin (log x)*dx` =


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


Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`


Integrate the following w.r.t.x : e2x sin x cos x


Evaluate the following.

`int x^2 e^4x`dx


Evaluate the following.

`int e^x (1/x - 1/x^2)`dx


Evaluate the following.

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


Choose the correct alternative from the following.

`int (1 - "x")^(-2) "dx"` = 


Evaluate: Find the primitive of `1/(1 + "e"^"x")`


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


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


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


Choose the correct alternative:

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


`int cot "x".log [log (sin "x")] "dx"` = ____________.


The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1)  dx` is


Find: `int e^x.sin2xdx`


`int 1/sqrt(x^2 - a^2)dx` = ______.


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


`int_0^1 x tan^-1 x  dx` = ______.


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


Evaluate :

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


Evaluate the following.

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


Solve the differential equation (x2 + y2) dx - 2xy dy = 0 by completing the following activity.

Solution: (x2 + y2) dx - 2xy dy = 0

∴ `dy/dx=(x^2+y^2)/(2xy)`                      ...(1)

Puty = vx

∴ `dy/dx=square`

∴ equation (1) becomes

`x(dv)/dx = square`

∴ `square  dv = dx/x`

On integrating, we get

`int(2v)/(1-v^2) dv =intdx/x`

∴ `-log|1-v^2|=log|x|+c_1`

∴ `log|x| + log|1-v^2|=logc       ...["where" - c_1 = log c]`

∴ x(1 - v2) = c

By putting the value of v, the general solution of the D.E. is `square`= cx


Solve the following

`int_0^1 e^(x^2) x^3 dx`


Evaluate:

`intcos^-1(sqrt(x))dx`


The value of `int e^x((1 + sinx)/(1 + cosx))dx` is ______.


Evaluate `int tan^-1x  dx`


Evaluate:

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


Evaluate the following:

`intx^3e^(x^2)dx` 


Evaluate:

`inte^x "cosec"  x(1 - cot x)dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×