मराठी

By using the properties of the definite integral, evaluate the integral: ∫0π2sinx-cosx1+sinxcosxdx - Mathematics

Advertisements
Advertisements

प्रश्न

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

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

बेरीज

उत्तर

`I = int_0^(pi/2) (sin x - cosx)/(1+sinx cos x) dx`     ....(i)

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

`I = int_0^(pi/2) (cosx-sinx)/(1+cosxsinx)dx`    .....(ii)

Adding (i) and (ii), we get :

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

`2I = int_0^(pi/2)(sinx-cosx+ cosx - sinx)/(1 +sinxcosx)    dx`

`2I = 0 ⇒I=0`

`⇒ int_0^(pi/2) (sinx-cosx)/(1+sinxcosx)  dx=0`  

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

APPEARS IN

एनसीईआरटी Mathematics [English] Class 12
पाठ 7 Integrals
Exercise 7.11 | Q 15 | पृष्ठ ३४७

संबंधित प्रश्‍न

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

`int_0^1 x(1-x)^n 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_0^(2x) cos^5 xdx`


\[\int_\pi^\frac{3\pi}{2} \sqrt{1 - \cos2x}dx\]

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


Evaluate`int (1)/(x(3+log x))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"`


Evaluate: `int_0^pi ("x"sin "x")/(1+ 3cos^2 "x") d"x"`.


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


`int_"a"^"b" "f"(x)  "d"x` = ______


`int_2^7 sqrt(x)/(sqrt(x) + sqrt(9 - x))  dx` = ______.


`int_1^2 1/(2x + 3)  dx` = ______


State whether the following statement is True or False:

`int_(-5)^5 x/(x^2 + 7)  "d"x` = 10


`int (cos x + x sin x)/(x(x + cos x))`dx = ?


The c.d.f, F(x) associated with p.d.f. f(x) = 3(1- 2x2). If 0 < x < 1 is k`(x - (2x^3)/"k")`, then value of k is ______.


`int_0^1 (1 - x/(1!) + x^2/(2!) - x^3/(3!) + ... "upto" ∞)` e2x dx = ?


`int_0^{pi/2} log(tanx)dx` = ______


`int_0^4 1/(1 + sqrtx)`dx = ______.


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


`int_0^{pi/2}((3sqrtsecx)/(3sqrtsecx + 3sqrt(cosecx)))dx` = ______ 


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


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


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


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} cos^2x  dx` = ______ 


`int_0^pi sin^2x.cos^2x  dx` = ______ 


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


`int_-1^1x^2/(1+x^2)  dx=` ______.


`int_(-pi/4)^(pi/4) 1/(1 - sinx) "d"x` = ______.


`int_0^1 "e"^(5logx) "d"x` = ______.


Find `int_0^(pi/4) sqrt(1 + sin 2x) "d"x`


`int_(-1)^1 (x^3 + |x| + 1)/(x^2 + 2|x| + 1) "d"x` is equal to ______.


If `int_0^1 "e"^"t"/(1 + "t") "dt"` = a, then `int_0^1 "e"^"t"/(1 + "t")^2 "dt"` is equal to ______.


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


Evaluate the following:

`int_0^(pi/2)  "dx"/(("a"^2 cos^2x + "b"^2 sin^2 x)^2` (Hint: Divide Numerator and Denominator by cos4x)


If `int (log "x")^2/"x" "dx" = (log "x")^"k"/"k" + "c"`, then the value of k is:


`int (dx)/(e^x + e^(-x))` is equal to ______.


Evaluate: `int_0^(π/2) 1/(1 + (tanx)^(2/3)) dx`


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


The integral `int_0^2||x - 1| -x|dx` is equal to ______.


Let `int_0^∞ (t^4dt)/(1 + t^2)^6 = (3π)/(64k)` then k is equal to ______.


Let f be continuous periodic function with period 3, such that `int_0^3f(x)dx` = 1. Then the value of `int_-4^8f(2x)dx` is ______.


With the usual notation `int_1^2 ([x^2] - [x]^2)dx` is equal to ______.


If `int_0^K dx/(2 + 18x^2) = π/24`, then the value of K is ______.


`int_0^(π/2)((root(n)(secx))/(root(n)(secx + root(n)("cosec"  x))))dx` is equal to ______.


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


`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: `int_0^π x/(1 + sinx)dx`.


Evaluate : `int_-1^1 log ((2 - x)/(2 + x))dx`.


Evaluate: `int_0^(π/4) log(1 + tanx)dx`.


Solve the following.

`int_1^3 x^2 logx  dx`


`int_1^2 x logx  dx`= ______


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_0^1x (1 - x)^5 dx`


Evaluate: `int_-1^1 x^17.cos^4x  dx`


Evaluate:

`int_0^1 |2x + 1|dx`


Solve the following.

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


Evaluate the following integral:

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


Evaluate the following integral:

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


Evaluate the following integral:

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


Solve the following.

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


Evaluate the following definite integral:

`int_-2^3(1)/(x + 5)  dx`


Evaluate the following definite intergral:

`int_1^3logx  dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×