Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
उत्तर
Let I = `int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
= `int_1^2 ("d"x)/sqrt(2x - x^2 - 2 + x)`
= `int_1^2 ("d"x)/sqrt(-x^2 + 3x - 2)`
= `int_1^2 ("d"x)/sqrt(-(x^2 - 3x + 2)`
= `int_1^2 ("d"x)/sqrt(-(x^2 - 3x + 9/4 - 9/4 + 2))` .....[Making perfect square]
= `int_1^2 ("d"x)/sqrt(-[(x - 3/2)^2 - 1/4])`
= `int_1^2 ("dx)/sqrt(1/4 - (x - 3/2)^2)`
= `int_1^2 ("d"x)/sqrt((1/2)^2 - (x - 3/2)^2)`
= `[sin^-1 ((x - 3/2)/(1/2))]_1^2`
= `[sin^-1 ((2x - 3)/1)]_1^2`
= `sin^-1 (4 - 3) - sin^-1 (2 - 3)`
= `sin^-1 (1) - sin^-1 (-1)`
= `sin^-1 (1) + sin^-1 (1)`
= `2 sin^-1 (1)`
= `2 xx pi/2`
= `pi`
Hence, I = `pi`.
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integral:
Write a value of
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate:
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`
