English

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

Advertisements
Advertisements

Question

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

`e^-x cos2x`

Sum
Advertisements

Solution

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`

  Is there an error in this question or solution?
Chapter 3: Indefinite Integration - Exercise 3.3 [Page 138]

APPEARS IN

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


Evaluate `int_0^(pi)e^2x.sin(pi/4+x)dx`


Integrate the function in x sin 3x.


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


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


Integrate the function in (x2 + 1) log x.


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


Integrate the function in e2x sin x.


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


`intx^2 e^(x^3) dx` equals: 


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


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


Evaluate the following:

`int sec^3x.dx`


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


Integrate the following functions w.r.t. x : `((1 + sin x)/(1 + cos x)).e^x`


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

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


Choose the correct options from the given alternatives :

`int cos -(3)/(7)x*sin -(11)/(7)x*dx` =


Choose the correct options from the given alternatives :

`int [sin (log x) + cos (log x)]*dx` =


Integrate the following w.r.t. x: `(1 + log x)^2/x`


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


Evaluate the following.

∫ x log x dx


Evaluate the following.

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


Evaluate the following.

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


Evaluate the following.

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


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


Evaluate:

∫ (log x)2 dx


`int 1/sqrt(2x^2 - 5)  "d"x`


`int sqrt(tanx) + sqrt(cotx)  "d"x`


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


`int"e"^(4x - 3) "d"x` = ______ + c


`int (x^2 + x - 6)/((x - 2)(x - 1))  "d"x` = x + ______ + c


Evaluate `int 1/(x log x)  "d"x`


`int "e"^x x/(x + 1)^2  "d"x`


Find `int_0^1 x(tan^-1x)  "d"x`


Evaluate the following:

`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`


Evaluate the following:

`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`


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


`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 ______.


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


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


`int1/sqrt(x^2 - a^2) dx` = ______


`intsqrt(1+x)  dx` = ______


`int (sin^-1 sqrt(x) + cos^-1 sqrt(x))dx` = ______.


Evaluate `int tan^-1x  dx`


If ∫(cot x – cosec2 x)ex dx = ex f(x) + c then f(x) will be ______.


Evaluate the following.

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×