मराठी

Integrate the function in x tan-1 x. - Mathematics

Advertisements
Advertisements

प्रश्न

Integrate the function in x tan-1 x.

बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.6 [पृष्ठ ३२७]

APPEARS IN

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

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Integrate the function in x sin−1 x.


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


Integrate the function in x sec2 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^3.tan^-1x.dx`


Evaluate the following : `int x.sin^2x.dx`


Evaluate the following : `int e^(2x).cos 3x.dx`


Integrate the following functions w.r.t. x : `e^(2x).sin3x`


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

`e^-x cos2x`


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


Integrate the following functions w.r.t. x : [2 + cot x – cosec2x]e 


Choose the correct options from the given alternatives :

`int tan(sin^-1 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 with respect to the respective variable : cos 3x cos 2x cos x


Integrate the following w.r.t.x : log (log x)+(log x)–2 


Evaluate the following.

`int "e"^"x" "x - 1"/("x + 1")^3` dx


Evaluate the following.

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


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


Choose the correct alternative from the following.

`int (1 - "x")^(-2) "dx"` = 


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


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


`int ("e"^xlog(sin"e"^x))/(tan"e"^x)  "d"x`


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


`int logx/(1 + logx)^2  "d"x`


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


Find `int_0^1 x(tan^-1x)  "d"x`


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


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


If `π/2` < x < π, then `intxsqrt((1 + cos2x)/2)dx` = ______.


`int e^x [(2 + sin 2x)/(1 + cos 2x)]dx` = ______.


`int_0^1 x tan^-1 x  dx` = ______.


`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`


Evaluate `int(3x-2)/((x+1)^2(x+3))  dx`


`int(xe^x)/((1+x)^2)  dx` = ______


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


Solve the following

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


Evaluate:

`int1/(x^2 + 25)dx`


Evaluate the following.

`int x^3 e^(x^2) dx` 


Evaluate the following.

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×