English

By using the properties of the definite integral, evaluate the integral: ∫0πx dx1+sinx

Advertisements
Advertisements

Question

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

`int_0^pi (x  dx)/(1+ sin x)`

Sum
Advertisements

Solution

Let `I = int_0^pi (x  dx)/ (1 + sin x)`

`I = int_0^pi (pi - x)/ (1 + sin (pi-x)) dx`           `...[∵ int_0^a f (x) dx = int_0^a f (a - x) dx]`

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

Adding (i) and (ii), we get

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

`= pi int_0^pi 1/ (1 +  sin x)  dx`

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

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

`= pi int_0^pi (sec^2 x - tan x sec x) dx`

`= pi [tan x - sec x]_0^pi`

= π [(tan π - sec π) - (tan0 - sec0)]

= π [(0 - (-1)) - (0 - 1)]

= 2π

Hence, I = π

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise 7.11 [Page 347]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 7 Integrals
Exercise 7.11 | Q 12 | Page 347

RELATED QUESTIONS

Prove that: `int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`


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

`int_0^(pi/2) (2log sin x - log sin 2x)dx`


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

`int_((-pi)/2)^(pi/2) sin^2 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 : \[\int(3x - 2) \sqrt{x^2 + x + 1}dx\] .


`int_0^2 e^x dx` = ______.


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


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


`int_0^1 (1 - x)^5`dx = ______.


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


If f(x) = |x - 2|, then `int_-2^3 f(x) dx` is ______


The value of `int_1^3 dx/(x(1 + x^2))` is ______ 


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


`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = ______ 


If `int_0^"k" "dx"/(2 + 32x^2) = pi/32,` then the value of k is ______.


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


The value of `int_2^7 (sqrtx)/(sqrt(9 - x) + sqrtx)dx` is ______ 


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


`int_(-1)^1 (x + x^3)/(9 - x^2)  "d"x` = ______.


`int_0^(pi/2) sqrt(1 - sin2x)  "d"x` is equal to ______.


`int_(-5)^5  x^7/(x^4 + 10)  dx` = ______.


Evaluate: `int_((-π)/2)^(π/2) (sin|x| + cos|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`


If `int_0^1(sqrt(2x) - sqrt(2x - x^2))dx = int_0^1(1 - sqrt(1 - y^2) - y^2/2)dy + int_1^2(2 - y^2/2)dy` + I then I equal.


If `int_(-a)^a(|x| + |x - 2|)dx` = 22, (a > 2) and [x] denotes the greatest integer ≤ x, then `int_a^(-a)(x + [x])dx` is equal to ______.


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


`int_0^1|3x - 1|dx` equals ______.


The value of the integral `int_0^sqrt(2)([sqrt(2 - x^2)] + 2x)dx` (where [.] denotes greatest integer function) is ______.


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


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


Evaluate the following integral:

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


`int_0^(2a)f(x)/(f(x)+f(2a-x))  dx` = ______


Evaluate the following definite integral:

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


Solve the following.

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


Evaluate:

`int_0^sqrt(2)[x^2]dx`


`int_(pi"/"11)^(9pi"/"22) (dx)/(1 + sqrttan x)` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×