Advertisements
Advertisements
प्रश्न
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Advertisements
उत्तर
Put x – α = t2.
Then β – x = β – (t2 +α)
= β – t2 – α
= – t2 – α + β
And dx = 2tdt.
Now I = `int (2"t dt")/sqrt("t"^2(beta - alpha - "t"^2))`
= `int (2"dt")/sqrt((beta - alpha - "t"^2))`
= `2 "dt"/sqrt("k"^2 - "t"^2)`, where k2 = β – α
= `2sin^-1 "t"/"k" + "C"`
= `2sin^-1 sqrt((x - alpha)/(beta - alpha)) + "C"`
APPEARS IN
संबंधित प्रश्न
If f is an integrable function, show that
Evaluate each of the following integral:
Solve each of the following integral:
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
