Advertisements
Advertisements
Question
Evaluate the following:
`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`
Advertisements
Solution
Let I = `int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`
= `int x^(3/2)/sqrt(("a"^(3/2))^2 - (x^(3/2))^2) "d"x`
Put `x^(3/2)` = t
⇒ `3/2 x^(1/2) "d"x` = dt
⇒ `x^(1/2) "d"x = 2/3 "dt"`
∴ I = `2/3 int "dt"/sqrt(("a"^(3/2))^2 - ("t")^2)`
= `2/3 sin^-1 "t"/("a"^(3/2)) + "C"`
= `2/3 sin^-1 ((x^(3/2))/("a"^(3/2))) + "C"`
Hence I = `2/3 sin^-1 (x/"a")^(3/2) + "C"`.
APPEARS IN
RELATED QUESTIONS
` ∫ cot^3 x "cosec"^2 x dx `
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integral:
Write a value of
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate:
`∫ (1)/(sin^2 x cos^2 x) dx`
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Evaluate the following:
`int x/(x^4 - 1) "d"x`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
