हिंदी

Integrate the following functions w.r.t. x: sin (log x) - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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

sin (log x)

योग
Advertisements

उत्तर

Le I  = `int sin (logx)x dx`

Put log x = t
∴ x = et
∴ dx = et dt

∴ I = `int sin t xx e^t dt`

= `int e^t sin t dt`

= `e^t int sin t dt - int [d/dt (e^t) int sin t dt] dt`

= `e^t (- cos t) - int e^t (- cos t) dt`

= `-e^t cos t + int e^t cos t dt`

= `- e^t cos t + e^t int cos t dt - int [d/dt (e^t) int cos t dt] dt`

= `- e^t cos t + e^t sin t - int e^t sin t dt`

∴ I = – et cos t + et sin t – I
∴ 2I = et (sin t – cos t)

∴ `I  = e^t/(2) (sin t - cos t) + c`

= `x/(2) [sin (logx) - cos (logx)] + c`.

shaalaa.com
  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 2.03 | पृष्ठ १३८

वीडियो 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`


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


Integrate the function in xlog x.


Integrate the function in x tan-1 x.


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


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


Find : 

`∫(log x)^2 dx`


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


Evaluate the following : `int cos sqrt(x).dx`


Evaluate the following : `int cos(root(3)(x)).dx`


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


Choose the correct options from the given alternatives :

`int (log (3x))/(xlog (9x))*dx` =


Choose the correct options from the given alternatives :

`int (sin^m x)/(cos^(m+2)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` =


Solve the following differential equation.

(x2 − yx2 ) dy + (y2 + xy2) dx = 0


Evaluate the following.

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


Choose the correct alternative from the following.

`int (("x"^3 + 3"x"^2 + 3"x" + 1))/("x + 1")^5  "dx"` = 


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


`int ["cosec"(logx)][1 - cot(logx)]  "d"x`


Evaluate `int 1/(4x^2 - 1)  "d"x`


`int 1/sqrt(x^2 - 8x - 20)  "d"x`


∫ log x · (log x + 2) dx = ?


`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.


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


Evaluate the following:

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


Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`


Evaluate the following:

`int_0^pi x log sin x "d"x`


The value of `int_0^(pi/2) log ((4 + 3 sin x)/(4 + 3 cos x))  dx` is


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 ______.


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


`int((4e^x - 25)/(2e^x - 5))dx = Ax + B log(2e^x - 5) + c`, then ______.


Find `int e^(cot^-1x) ((1 - x + x^2)/(1 + x^2))dx`.


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


Evaluate :

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


`intsqrt(1+x)  dx` = ______


Evaluate the following.

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


Evaluate:

`int((1 + sinx)/(1 + cosx))e^x dx`


Evaluate:

`int (logx)^2 dx`


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


Complete the following activity:

`int_0^2 dx/(4 + x - x^2) `

= `int_0^2 dx/(-x^2 + square + square)`

= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`

= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`

= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`


Evaluate the following.

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


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


Evaluate the following. 

`int x sqrt(1 + x^2)  dx`  


Evaluate the following.

`int x^3 e^(x^2) dx` 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×