English

Integrate the function in tan-1 x.

Advertisements
Advertisements

Question

Integrate the function in tan-1 x.

Sum
Advertisements

Solution

Let `I = int tan^-1 x  dx`

`= int tan^-1 x. 1  dx`

Put `u = tan^-1 x, v = 1` 

`int uv  dx = u int v  dx - int ((du)/dx int v  dx)  dx`

`I= int tan^-1 x. 1`

`(tan^-1 x) int 1  dx - (d/dx (tan^-1 x) int dx) dx`

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

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

Put 1 + x2 = t, and dx = dt

`= x tan^-1 x - 1/2 int dt/t`

`= x tan^-1 x - 1/2  log t + C`

`= x tan^1 - 1/2  log (1 + x^2) + C`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise 7.6 [Page 327]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 7 Integrals
Exercise 7.6 | Q 13 | Page 327

RELATED QUESTIONS

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 log 2x.


Integrate the function in xlog x.


Integrate the function in x tan-1 x.


Integrate the function in e2x sin x.


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


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


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


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


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

sin (log x)


Integrate the following functions w.r.t. x: `sqrt(x^2 + 2x + 5)`.


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


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 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 : cot–1 (1 – x + x2)


Integrate the following w.r.t.x : `(1)/(x^3 sqrt(x^2 - 1)`


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


Solve the following differential equation.

(x2 − yx2 ) dy + (y2 + xy2) dx = 0


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


Evaluate:

∫ (log x)2 dx


`int (cos2x)/(sin^2x cos^2x)  "d"x`


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


`int (x^2 + x - 6)/((x - 2)(x - 1))  "d"x` = x + ______ + c


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


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


Evaluate the following:

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


State whether the following statement is true or false.

If `int (4e^x - 25)/(2e^x - 5)` dx = Ax – 3 log |2ex – 5| + c, where c is the constant of integration, then A = 5.


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


Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`


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 ______.


`int(1-x)^-2 dx` = ______


`int1/sqrt(x^2 - a^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


Evaluate:

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


Evaluate the following.

`intx^3  e^(x^2) dx`


Evaluate the following.

`intx^3/sqrt(1+x^4)dx`


Evaluate the following.

`intx^3/sqrt(1+x^4)  dx`


The value of `inta^x.e^x dx` equals


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×