मराठी

Given that ∞ ∫ 0 X 2 ( X 2 + a 2 ) ( X 2 + B 2 ) ( X 2 + C 2 ) D X = π 2 ( a + B ) ( B + C ) ( C + a ) , the Value of ∞ ∫ 0 D X ( X 2 + 4 ) ( X 2 + 9 ) , - Mathematics

Advertisements
Advertisements

प्रश्न

Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]

पर्याय

  • \[\frac{\pi}{60}\]
  • \[\frac{\pi}{20}\]
  • \[\frac{\pi}{40}\]
  • \[\frac{\pi}{80}\]
MCQ

उत्तर

\[\frac{\pi}{60}\]

 

\[ \int_0^\infty \frac{1}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)} d x\]
\[ = \frac{1}{5} \int_0^\infty \frac{1}{\left( x^2 + 4 \right)} - \frac{1}{\left( x^2 + 9 \right)}dx\]
\[ = \frac{1}{5} \left[ \frac{1}{2} \tan^{- 1} \frac{x}{2} - \frac{1}{3} \tan^{- 1} \frac{x}{3} \right]_0^\infty \]
\[ = \frac{1}{5}\left[ \frac{1}{2} \times \frac{\pi}{2} - \frac{1}{3} \times \frac{\pi}{2} \right]\]
\[ = \frac{1}{5} \times \frac{\pi}{12}\]
\[ = \frac{\pi}{60}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - MCQ [पृष्ठ ११८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
MCQ | Q 13 | पृष्ठ ११८

संबंधित प्रश्‍न

\[\int\limits_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx\]

\[\int\limits_{- 1}^1 \frac{1}{1 + x^2} dx\]

\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\]

\[\int\limits_0^{\pi/2} \sin x \sin 2x\ dx\]

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_1^e \frac{e^x}{x} \left( 1 + x \log x \right) dx\]

\[\int\limits_{\pi/2}^\pi e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx\]

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

\[\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0\]

\[\int\limits_0^\pi \frac{1}{3 + 2 \sin x + \cos x} dx\]

\[\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt\]

\[\int\limits_0^{\pi/2} \cos^5 x\ dx\]

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]

\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

\[\int\limits_0^{\pi/2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} dx\]

\[\int\limits_0^\pi \frac{x \sin x}{1 + \sin x} dx\]

\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]

\[\int\limits_2^3 \frac{1}{x}dx\]

If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.

 

 


Evaluate : 

\[\int\limits_2^3 3^x dx .\]

\[\int\limits_0^{15} \left[ x \right] dx .\]

\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.  

 

The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .


Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


Using second fundamental theorem, evaluate the following:

`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`


Evaluate the following:

`int_0^oo "e"^(- x/2) x^5  "d"x`


Choose the correct alternative:

Γ(n) is


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


Find `int x^2/(x^4 + 3x^2 + 2) "d"x`


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×