हिंदी

Evaluate ∫π0 e^2 x.sin(π/4+x) dx - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

Let `I==int_0^pie^(2x)sin(pi/2+x)dx`

Integrating by parts, we get

` I=1/2[e^(2x)sin(pi/4+x)_0^pi]-1/2int_0^pie^(2x)cos(pi/4+x)dx`

 Now, integrating the second term by parts, we get

` =>I=1/2[e^(2x)sin(pi/4+x)_0^pi]-1/2{[1/2e^(2x)cos(pi/4+x)_0^pi]+1/2int_0^pi e^(2x)sin(pi/4+x)dx}`

=>`I=1/2[e^(2x)sin(pi/4+x)_0^pi]-1/4[e^(2x)cos(pi/4+x)_0^pi]-1/4I`

`=>5/4I=1/2[e^(2x)sin(pi+pi/4)-sin(pi/4)]-1/4[e^(2x)cos(pi+pi/4)-cos(pi/4)]`

`=>5/4I=1/2 |__-e^(2x)xx1/sqrt2-1/sqrt2__|-1/4|__-e^(2pi)xx1/sqrt2-1/sqrt2__|`

`=>5/4I==1/(2sqrt2)e^(2pi)-1/(2sqrt2)+1/(4sqrt2)e^(2pi)+1/(4sqrt2)`

`=>I=-1/(5sqrt2)(e^(2pi)+1)`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2015-2016 (March) Delhi Set 1

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

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


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 cos^(-1) x)/sqrt(1-x^2)`.


Integrate the function in tan-1 x.


Integrate the function in x (log x)2.


Evaluate the following:

`int sec^3x.dx`


Evaluate the following : `int x.sin^2x.dx`


Evaluate the following : `int e^(2x).cos 3x.dx`


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


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

`e^-x cos2x`


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


Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`


Choose the correct options from the given alternatives :

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


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


Integrate the following w.r.t.x : sec4x cosec2x


Evaluate: `int e^x/sqrt(e^(2x) + 4e^x + 13)` dx


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


`int cot "x".log [log (sin "x")] "dx"` = ____________.


The value of `int "e"^(5x) (1/x - 1/(5x^2))  "d"x` is ______.


Find `int_0^1 x(tan^-1x)  "d"x`


Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`


`int(logx)^2dx` equals ______.


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


`intsqrt(1+x)  dx` = ______


`int(xe^x)/((1+x)^2)  dx` = ______


Solve the following

`int_0^1 e^(x^2) x^3 dx`


Evaluate the following.

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


Evaluate the following. 

`int x sqrt(1 + x^2)  dx`  


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×