मराठी

By using the properties of the definite integral, evaluate the integral: ∫0π2 cos5 xdxsin5x+cos5x

Advertisements
Advertisements

प्रश्न

By using the properties of the definite integral, evaluate the integral:

`int_0^(pi/2)  (cos^5  xdx)/(sin^5 x + cos^5 x)`

बेरीज
Advertisements

उत्तर

Let `I = int_0^(pi/2) (cos^5 x)/ (sin^5 x + cos ^5 x)  dx`     ....(i)

Also, `I = int_0^(pi/2) (cos^5 (pi/2 - x))/(sin^5 (pi/2 - x) + cos^5 (pi/2 - x)) dx`

`[∵ int_0^a f (x) dx = int_0^a f (a - x) dx]`

`= int_0^(pi/2) (sin^5 x)/ (cos^5x + sin^5 x)  dx`           ....(ii)

Adding (i) and (ii), we have

`2 I = int_0^(pi/2) (cos^5x)/(cos^5x + sin^5 x)  dx + int_0^(pi/2) (sin^5x)/ (cos^5 x + sin^5 x)  dx`

`= int_0^(pi/2) (cos^5 x + sin^5 x)/ (cos^5 x + sin^5 x) dx`

`= int_0^(pi/2) 1 dx = [x]_0^(pi/2) = pi/2`

Hence, `I = pi/4`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.11 [पृष्ठ ३४७]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.11 | Q 4 | पृष्ठ ३४७

संबंधित प्रश्‍न

If `int_0^alpha3x^2dx=8` then the value of α is :

(a) 0

(b) -2

(c) 2 

(d) ±2


Evaluate : `int e^x[(sqrt(1-x^2)sin^-1x+1)/(sqrt(1-x^2))]dx`


By using the properties of the definite integral, evaluate the integral:

`int_0^2 xsqrt(2 -x)dx`


By using the properties of the definite integral, evaluate the integral:

`int_0^pi (x  dx)/(1+ sin x)`


By using the properties of the definite integral, evaluate the integral:

`int_(pi/2)^(pi/2) sin^7 x dx`


By using the properties of the definite integral, evaluate the integral:

`int_0^(pi/2) (sin x - cos x)/(1+sinx cos x) dx`


By using the properties of the definite integral, evaluate the integral:

`int_0^4 |x - 1| dx`


Show that `int_0^a f(x)g (x)dx = 2 int_0^a f(x) dx`  if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4.


Evaluate the definite integrals `int_0^pi (x tan x)/(sec x + tan x)dx`


Evaluate `int_0^(pi/2) cos^2x/(1+ sinx cosx) dx`


\[\int\limits_0^k \frac{1}{2 + 8 x^2} dx = \frac{\pi}{16},\] find the value of k.


Evaluate : \[\int(3x - 2) \sqrt{x^2 + x + 1}dx\] .


Evaluate : `int 1/("x" [("log x")^2 + 4])  "dx"`


The total revenue R = 720 - 3x2 where x is number of items sold. Find x for which total  revenue R is increasing.


Evaluate :  ∫ log (1 + x2) dx


Prove that `int_0^"a" "f" ("x") "dx" = int_0^"a" "f" ("a" - "x") "d x",` hence evaluate `int_0^pi ("x" sin "x")/(1 + cos^2 "x") "dx"`


Choose the correct alternative:

`int_(-9)^9 x^3/(4 - x^2)  "d"x` =


`int_2^3 x/(x^2 - 1)` dx = ______


`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = ______.


`int_0^{pi/4} (sin2x)/(sin^4x + cos^4x)dx` = ____________


f(x) =  `{:{(x^3/k;       0 ≤ x ≤ 2), (0;     "otherwise"):}` is a p.d.f. of X. The value of k is ______


`int_0^{pi/2} (cos2x)/(cosx + sinx)dx` = ______


`int_-1^1x^2/(1+x^2)  dx=` ______.


`int_0^(pi/2) 1/(1 + cos^3x) "d"x` = ______.


Evaluate the following:

`int_(-pi/4)^(pi/4) log|sinx + cosx|"d"x`


`int_0^(pi/2)  cos x "e"^(sinx)  "d"x` is equal to ______.


The value of `int_0^1 tan^-1 ((2x - 1)/(1 + x - x^2))  dx` is


If `int_a^b x^3 dx` = 0, then `(x^4/square)_a^b` = 0

⇒ `1/4 (square - square)` = 0

⇒ b4 – `square` = 0

⇒ (b2 – a2)(`square` + `square`) = 0

⇒ b2 – `square` = 0 as a2 + b2 ≠ 0

⇒ b = ± `square`


`int_a^b f(x)dx = int_a^b f(x - a - b)dx`.


If `β + 2int_0^1x^2e^(-x^2)dx = int_0^1e^(-x^2)dx`, then the value of β is ______.


`int_-1^1 |x - 2|/(x - 2) dx`, x ≠ 2 is equal to ______.


The value of `int_0^(π/4) (sin 2x)dx` is ______.


Evaluate the following definite integral:

`int_-2^3 1/(x + 5) dx`


Solve the following.

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


Evaluate the following integral:

`int_-9^9x^3/(4-x^2)dx`


Solve the following.

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


\[\int_{-2}^{2}\left|x^{2}-x-2\right|\mathrm{d}x=\]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×