मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Evaluate the following : ∫x2tan-1x.dx

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

Let I = `int x^2 tan^-1 x.dx`

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

= `(tan^-1x) int x^2.dx - int[{d/dx(tan^-1x) int x^2.dx}].dx`

= `(tan^-1 x)(x^3/3) - int (1/(1 + x^2))(x^3/3).dx`

= `x3/(3) tan^-1x - (1)/(3) (x(x^2 + 1) - x)/(x^2 + 1).dx`

= `x^3/(3) tan^-1x - (1)/(3)[int{x - x/(x^2 + 1)}.dx]`

= `x^3/(3) tan^-1x - (1)/(3)[int x.dx - (1)/(2) int(2x)/(x^2 + 1).dx]`

= `x^3/(3)tan^-1x - (1)/(3) [x^2/(2) - (1)/(2)log|x^2 + 1|] + c`

...`[because d/dx(x^2 + 1) = 2x and int (f'(x))/f(x) dx = log|f(x)| + c]`

= `x^3/(3)tan^-1x - x^2/(6) + (1)/(6) log|x^2 + 1| + c`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Indefinite Integration - Exercise 3.3 [पृष्ठ १३७]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Indefinite Integration
Exercise 3.3 | Q 1.04 | पृष्ठ १३७

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

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`


If `int_(-pi/2)^(pi/2)sin^4x/(sin^4x+cos^4x)dx`, then the value of I is:

(A) 0

(B) π

(C) π/2

(D) π/4


`int1/xlogxdx=...............`

(A)log(log x)+ c

(B) 1/2 (logx )2+c

(C) 2log x + c

(D) log x + c


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


Integrate the function in ex (sinx + cosx).


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


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


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.sin^2x.dx`


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


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


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


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

`e^-x cos2x`


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


Integrate the following functions w.r.t. x : `e^(sin^-1x)*[(x + sqrt(1 - x^2))/sqrt(1 - x^2)]`


Choose the correct options from the given alternatives :

`int (log (3x))/(xlog (9x))*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 


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


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


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


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


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


Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`


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


`int 1/sqrt(x^2 - 9) dx` = ______.


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


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


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


If `int (f(x))/(log(sin x))dx` = log[log sin x] + c, then f(x) is equal to ______.


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


Find `int e^x ((1 - sinx)/(1 - cosx))dx`.


Evaluate :

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


The integrating factor of `ylogy.dx/dy+x-logy=0` is ______.


`int(xe^x)/((1+x)^2)  dx` = ______


Evaluate the following.

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


Evaluate:

`int (logx)^2 dx`


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`


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


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


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


Evaluate the following. 

`int x sqrt(1 + x^2)  dx`  


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×