हिंदी

Integrate the function in x tan-1 x.

Advertisements
Advertisements

प्रश्न

Integrate the function in x tan-1 x.

योग
Advertisements

उत्तर

Let `I = int x tan^-1 x dx`

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

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

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

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

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

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

`= x^2/2 tan^-1 x - 1/2 x + 1/2 tan^-1 x + C`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Exercise 7.6 [पृष्ठ ३२७]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.6 | Q 8 | पृष्ठ ३२७

वीडियो ट्यूटोरियलVIEW ALL [2]

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`


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 sin−1 x.


Integrate the function in x cos-1 x.


Integrate the function in e2x sin x.


`intx^2 e^(x^3) dx` equals: 


Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`


Integrate the following functions w.r.t. x : `x^2 .sqrt(a^2 - x^6)`


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


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

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


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 tan(sin^-1 x)*dx` =


Integrate the following with respect to the respective variable : cos 3x cos 2x cos x


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


Evaluate the following.

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


Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx


Evaluate: `int "dx"/sqrt(4"x"^2 - 5)`


`int 1/(4x + 5x^(-11))  "d"x`


`int (cos2x)/(sin^2x cos^2x)  "d"x`


Choose the correct alternative:

`int ("d"x)/((x - 8)(x + 7))` =


`int(x + 1/x)^3 dx` = ______.


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


`int "e"^x x/(x + 1)^2  "d"x`


Evaluate the following:

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


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


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


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


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


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


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


`int logx  dx = x(1+logx)+c`


Evaluate:

`int (logx)^2 dx`


If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x)


Evaluate:

`int x^2 cos x  dx`


Evaluate the following.

`int x^3 e^(x^2) dx` 


The value of `int (x sin^-1)/(sqrt(1 - x^2)) dx` is equal to:


`∫ sin^(−1)` xdx is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×