Advertisements
Advertisements
Question
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 \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
Options
- \[\frac{\pi}{60}\]
- \[\frac{\pi}{20}\]
- \[\frac{\pi}{40}\]
- \[\frac{\pi}{80}\]
Advertisements
Solution
\[\frac{\pi}{60}\]
\[ \int_0^\infty \frac{1}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)} d x\]
\[ = \frac{1}{5} \int_0^\infty \frac{1}{\left( x^2 + 4 \right)} - \frac{1}{\left( x^2 + 9 \right)}dx\]
\[ = \frac{1}{5} \left[ \frac{1}{2} \tan^{- 1} \frac{x}{2} - \frac{1}{3} \tan^{- 1} \frac{x}{3} \right]_0^\infty \]
\[ = \frac{1}{5}\left[ \frac{1}{2} \times \frac{\pi}{2} - \frac{1}{3} \times \frac{\pi}{2} \right]\]
\[ = \frac{1}{5} \times \frac{\pi}{12}\]
\[ = \frac{\pi}{60}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
