Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\infty \frac{1}{a^2 + b^2 x^2} d x\ . Then, \]
\[I = \frac{1}{a^2} \int_0^\infty \frac{1}{1 + \frac{b^2 x^2}{a^2}} d x\]
\[ \Rightarrow I = \frac{1}{a^2} \int_0^\infty \frac{1}{1 + \left( \frac{bx}{a} \right)^2} d x\]
\[ \Rightarrow I = \frac{a}{b a^2} \left[ \tan^{- 1} \left( \frac{bx}{a} \right) \right]_0^\infty \]
\[ \Rightarrow I = \frac{1}{ab}\left( \tan^{- 1} \infty - \tan^{- 1} 0 \right)\]
\[ \Rightarrow I = \frac{\pi}{2ab}\]
APPEARS IN
संबंधित प्रश्न
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
If f(x) is a continuous function defined on [−a, a], then prove that
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
`int_0^(2a)f(x)dx`
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
