Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Advertisements
उत्तर
Let I = `int ("d"x)/sqrt(16 - 9x^2)`
= `1/3 int ("d"x)/sqrt(16/9 - x^2)`
= `1/3 int ("d"x)/sqrt((4/3)^2 - x^2)`
= `1/3 sin^-1 x/(4/3) + "C"` ....`[because int ("d"x)/sqrt("a"^2 - x^2) = sin^-1 x/"a" + "C"]`
∴ I = `1/3 sin^-1 (3x)/4 + "C"`
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_0^3dx/(9+x^2)`
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate:\[\int\frac{x^2}{1 + x^3} \text{ dx }\] .
Evaluate:
Evaluate:\[\int\frac{\cos \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
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 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 sqrt(5 - 2x + x^2) "d"x`
Evaluate the following:
`int x/(x^4 - 1) "d"x`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
