English

Choose the correct options from the given alternatives : ∫sinmxcosm+2x⋅dx = - Mathematics and Statistics

Advertisements
Advertisements

Question

Choose the correct options from the given alternatives :

`int (sin^m x)/(cos^(m+2)x)*dx` = 

Options

  • `(tan^(m+1)x)/(m + 1) + c`

  • (m + 2)tanm+1 x + c

  • `tan^mx/m + c`

  • (m + 1)tanm+1 x + c

MCQ
Advertisements

Solution

`(tan^(m+1)x)/(m + 1) + c`

shaalaa.com
  Is there an error in this question or solution?
Chapter 3: Indefinite Integration - Miscellaneous Exercise 3 [Page 148]

APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 3 Indefinite Integration
Miscellaneous Exercise 3 | Q 1.04 | Page 148

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

`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 x sin 3x.


Integrate the function in x log x.


Integrate the function in x log 2x.


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


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


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


Find : 

`∫(log x)^2 dx`


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


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


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


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


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


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


If f(x) = `sin^-1x/sqrt(1 - x^2), "g"(x) = e^(sin^-1x)`, then `int f(x)*"g"(x)*dx` = ______.


Integrate the following with respect to the respective variable : `(3 - 2sinx)/(cos^2x)`


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


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 : sec4x cosec2x


Evaluate the following.

`int [1/(log "x") - 1/(log "x")^2]` dx


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


Evaluate:

∫ (log x)2 dx


`int sin4x cos3x  "d"x`


`int 1/x  "d"x` = ______ + c


∫ log x · (log x + 2) dx = ?


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


Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`


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


`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`


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


`int1/sqrt(x^2 - a^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


Evaluate:

`int e^(ax)*cos(bx + c)dx`


Evaluate:

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


If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv  dx - int(d/dx u)(intv  dx)dx`. Hence evaluate: `intx cos x  dx`


Evaluate the following.

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


Evaluate the following.

`intx^3/sqrt(1+x^4)  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`  


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×