हिंदी

If ∫π/2 −π/2 sin^4 x/ (sin^4 x+cos^4 x)dx, then the value of I is:

Advertisements
Advertisements

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

Advertisements

उत्तर

(C)

`I= int_(-pi/2)^(pi/2)(sin^4x)/(sin^4x+cos^4x)dx`

`Let f(x)=(sin^4x)/(sin^4x+cos^4x)dx`

`f(x)=(sin^4(-x))/(sin^4(-x)+cos^4(-x_)dx`

`f(x)=(sin^4x)/(sin^4x+cos^4x)dx`

=f(x)

f(x) is an even function.

`I=2 int_(0)^(pi/2)(sin^4x)/(sin^4x+cos^4x)dx............(i)`

`I=2 int_(0)^(pi/2)(sin^4(pi/2-x))/(sin^4(pi/2-x)+cos^4(pi/2-x))dx`

`I=2 int_(0)^(pi/2)(cos^4x)/(cos^4x+sin^4x)dx......(ii)`

Adding (i) and (ii), we get

`2I=2 int_(0)^(pi/2)dx`

`I=[x]_0^(pi/2)`

`I=pi/2`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2014-2015 (October)

APPEARS IN

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 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 log x.


Integrate the function in x log 2x.


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


Evaluate the following:

`int x tan^-1 x . dx`


Evaluate the following:

`int sec^3x.dx`


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


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


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


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

sin (log x)


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


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


Choose the correct options from the given alternatives :

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


Integrate the following w.r.t.x : log (log x)+(log x)–2 


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


Evaluate the following.

`int e^x (1/x - 1/x^2)`dx


Evaluate the following.

`int "e"^"x" "x - 1"/("x + 1")^3` dx


Choose the correct alternative from the following.

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


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


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


Choose the correct alternative:

`int ("d"x)/((x - 8)(x + 7))` =


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


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


Evaluate `int 1/(x log x)  "d"x`


`int "e"^x x/(x + 1)^2  "d"x`


`int 1/sqrt(x^2 - 8x - 20)  "d"x`


∫ log x · (log x + 2) dx = ?


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


The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1)  dx` is


Find: `int e^x.sin2xdx`


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 ______.


`intsqrt(1+x)  dx` = ______


Solution of the equation `xdy/dx=y log y` 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))`


If f'(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x)


`∫ sin^(−1)` xdx is equal to ______.


Under the LIATE rule, which type of function has priority \(L\)?


For the special integral \[\int e^x\left[f(x)+f'(x)\right]dx,\] what is the antiderivative?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×