मराठी

∫ 3 π 2 π √ 1 − Cos 2 X D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int_\pi^\frac{3\pi}{2} \sqrt{1 - \cos2x}dx\]
\[= \int_\pi^\frac{3\pi}{2} \sqrt{2 \sin^2 x}dx\]
\[ = \sqrt{2} \int_\pi^\frac{3\pi}{2} \left| \sin x \right|dx\]
\[ = - \sqrt{2} \int_\pi^\frac{3\pi}{2} \sin x\ dx .................\left( \sin x < 0 for\ \pi \leq x \leq 2\pi \right)\]

\[= - \sqrt{2}\left( - \cos x \right) |_\pi^\frac{3\pi}{2} \]
\[ = \sqrt{2}\left( \cos\frac{3\pi}{2} - cos\pi \right)\]
\[ = \sqrt{2} \left[ 0 - \left( - 1 \right) \right]\]
\[ = \sqrt{2} \times 1\]
\[ = \sqrt{2}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.1 [पृष्ठ १८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.1 | Q 62 | पृष्ठ १८

संबंधित प्रश्‍न

 
 

Evaluate : `intlogx/(1+logx)^2dx`

 
 

 
 

Evaluate `int_(-2)^2x^2/(1+5^x)dx`

 
 

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

`int_0^(pi/4) log (1+ tan x) 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/2) (sin x - cos x)/(1+sinx cos x) 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 `int_0^(pi/2) cos^2x/(1+ sinx cosx) dx`


If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that

\[\int_a^b xf\left( x \right)dx = \left( \frac{a + b}{2} \right) \int_a^b f\left( x \right)dx\]

Evaluate :  ∫ log (1 + x2) dx


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


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


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


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


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


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


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


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


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


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


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 `lim_("n"→∞)(int_(1/("n"+1))^(1/"n") tan^-1("n"x)"d"x)/(int_(1/("n"+1))^(1/"n") sin^-1("n"x)"d"x) = "p"/"q"`, (where p and q are coprime), then (p + q) is ______.


`int_((-π)/2)^(π/2) log((2 - sinx)/(2 + sinx))` is equal to ______.


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


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


Evaluate the following limit :

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


Evaluate the following integral:

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


`int_1^2 x logx  dx`= ______


Evaluate the following integral:

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


Solve the following.

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


Evaluate the following integral:

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


Evaluate the following definite integral:

`int_-2^3(1)/(x + 5)  dx`


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


`int_0^(pi/4) (cos^2 x)/(cos^2 x + 4 sin^2 x) dx` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×