हिंदी

Evaluate d∫0π2tan7xcot7x+tan7xdx - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate `int_0^(pi/2) (tan^7x)/(cot^7x + tan^7x) "d"x`

योग
Advertisements

उत्तर

We have I = `int_0^(pi/2) (tan^7x)/(cot^7x + tan^7x) "d"x`  ....(1)

= `int_0^(pi/2) (tan^7(pi/2 - x))/(cot^7(pi/2 - x) + tan^7(pi/2 - x)) "d"x` ......By (p4)

= `int_0^(pi/2) (cot^7 (x) "d"x)/(cot^7x "d"x + tan^7x)`  .....(2)

Adding (1) and (2), we get

2I = `int_0^(pi/2) ((tan^7x + cot^7x)/(tan^7x + cot^7x))"d"x`

= `int_0^(pi/2) "d"x` which gives I = `pi/4`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Solved Examples [पृष्ठ १५१]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 7 Integrals
Solved Examples | Q 10 | पृष्ठ १५१

संबंधित प्रश्न

 
 

Evaluate `int_(-2)^2x^2/(1+5^x)dx`

 
 

If `int_0^alpha(3x^2+2x+1)dx=14` then `alpha=`

(A) 1

(B) 2

(C) –1

(D) –2


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

`int_0^(pi/2) cos^2 x dx`


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

`int_0^(pi/2) sin^(3/2)x/(sin^(3/2)x + cos^(3/2) x) dx`


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`


Evaluate`int (1)/(x(3+log x))dx` 


Evaluate :  ∫ log (1 + x2) dx


Using properties of definite integrals, evaluate 

`int_0^(π/2)  sqrt(sin x )/ (sqrtsin x + sqrtcos x)dx`


Find : `int_  (2"x"+1)/(("x"^2+1)("x"^2+4))d"x"`.


Evaluate the following integrals : `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7 - x))*dx`


Evaluate the following integral:

`int_0^1 x(1 - x)^5 *dx`


`int_0^{pi/2} xsinx dx` = ______


`int_"a"^"b" sqrtx/(sqrtx + sqrt("a" + "b" - x)) "dx"` = ______.


If `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`, then K = ______.


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


`int_-2^1 dx/(x^2 + 4x + 13)` = ______


`int_(-1)^1 log ((2 - x)/(2 + x)) "dx" = ?`


`int_(pi/4)^(pi/2) sqrt(1-sin 2x)  dx =` ______.


Show that `int_0^(pi/2) (sin^2x)/(sinx + cosx) = 1/sqrt(2) log (sqrt(2) + 1)`


Evaluate the following:

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


If `int_0^"a" 1/(1 + 4x^2) "d"x = pi/8`, then a = ______.


`int_0^(2"a") "f"("x") "dx" = int_0^"a" "f"("x") "dx" + int_0^"a" "f"("k" - "x") "dx"`, then the value of k is:


Evaluate: `int_(pi/6)^(pi/3) (dx)/(1 + sqrt(tanx)`


Evaluate: `int_(-1)^3 |x^3 - x|dx`


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`


Let a be a positive real number such that `int_0^ae^(x-[x])dx` = 10e – 9 where [x] is the greatest integer less than or equal to x. Then, a is equal to ______.


The value of `int_((-1)/sqrt(2))^(1/sqrt(2)) (((x + 1)/(x - 1))^2 + ((x - 1)/(x + 1))^2 - 2)^(1/2)`dx is ______.


If f(x) = `{{:(x^2",", "where"  0 ≤ x < 1),(sqrt(x)",", "when"  1 ≤ x < 2):}`, then `int_0^2f(x)dx` equals ______.


`int_0^(π/4) x. sec^2 x  dx` = ______.


`int_-1^1 (17x^5 - x^4 + 29x^3 - 31x + 1)/(x^2 + 1) dx` is equal to ______.


Evaluate `int_-1^1 |x^4 - x|dx`.


Evaluate the following limit :

`lim_("x"->3)[sqrt("x"+6)/"x"]`


Evaluate `int_0^3root3(x+4)/(root3(x+4)+root3(7-x))  dx`


Evaluate the following definite integral:

`int_1^3 log x  dx`


Evaluate the following integral:

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


Evaluate the following integral:

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


Evaluate the following integrals:

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


Evaluate the following integral:

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


Solve the following.

`int_2^3x/((x+2)(x+3))dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×