हिंदी

Integrate the following functions w.r.t. x : e-xcos2x - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

`e^-x cos2x`

योग
Advertisements

उत्तर

Let I = `int e^-x cos 2x.dx`

∫ uv dx = u ∫ v dx – ∫ (u' ∫ v dx) dx.

I = `cos 2x int e^-x  dx  – int int e^(-x). d/dx cos 2x. dx`

I = `cos 2x. (e^-x)/(d/dx (-x)) – int(e^-x)/(d/dx (- x)). (- sin 2x. d/dx 2x) dx`

I = `- cos 2x. e^-x  – int (- e^(-x)) . (- 2sin 2x) dx`

I = `- cos 2x. e^-x  –  2 int e^(-x). sin 2x  dx`

I = `- cos 2x. e^-x - 2 [sin 2x. int e^-x dx - int int e^(-x) dx. d/dx sin 2x. dx]`

I = `- cos 2x. e^-x - 2 sin 2x. (e^-x)/(- 1) + 2 int  (e^-x)/(- 1). cos 2x. 2. dx`

I = `- cos 2x. e^-x + 2 sin 2x. (e^-x) - 2 int  2. e^(-x). cos 2x.dx`

I = `- cos 2x. e^-x + 2 sin 2x. (e^-x) - 4 int  e^(-x). cos 2x.dx`

I = `- cos 2x. e^-x + 2 sin 2x. (e^-x) - 4I`

I + 4I = `- cos 2x. e^-x + 2 sin 2x. (e^-x)`

5I = `e^-x (2. sin 2x - cos 2x)`

I = `e^-x/5  (2. sin 2x - cos 2x) + C`

shaalaa.com

Notes

Let I = `int e^-x cos 2x.dx`

∫ uv dx = u ∫ v dx – ∫ (u' ∫ v dx) dx.

I = `cos 2x int e^-x  dx  – int int e^(-x). d/dx cos 2x. dx`

  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Exercise 3.3 [पृष्ठ १३८]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Exercise 3.3 | Q 2.02 | पृष्ठ १३८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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 sin 3x.


Integrate the function in x tan-1 x.


Integrate the function in x (log x)2.


Integrate the function in e2x sin x.


`int e^x sec x (1 +   tan x) dx` equals:


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^2.log x.dx`


Evaluate the following:

`int sec^3x.dx`


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


Integrate the following functions w.r.t. x : `sqrt(5x^2 + 3)`


Integrate the following functions w.r.t. x : `xsqrt(5 - 4x - x^2)`


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


Choose the correct options from the given alternatives :

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


Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`


Solve the following differential equation.

(x2 − yx2 ) dy + (y2 + xy2) dx = 0


Evaluate the following.

∫ x log x dx


Evaluate the following.

`int (log "x")/(1 + log "x")^2` dx


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


`int (sin(x - "a"))/(cos (x + "b"))  "d"x`


`int (cos2x)/(sin^2x cos^2x)  "d"x`


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


Evaluate `int 1/(x(x - 1))  "d"x`


Evaluate `int 1/(x log x)  "d"x`


Evaluate `int 1/(4x^2 - 1)  "d"x`


∫ log x · (log x + 2) dx = ?


`int "dx"/(sin(x - "a")sin(x - "b"))` is equal to ______.


Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.


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 ______.


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


`int_0^1 x tan^-1 x  dx` = ______.


Evaluate :

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


`inte^(xloga).e^x dx` is ______


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


`int(xe^x)/((1+x)^2)  dx` = ______


`int(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`


Evaluate the following.

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


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 sqrt(1 + x^2)  dx`  


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×