Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)
उत्तर
Let I = `int ("d"x)/(xsqrt(x^4 - 1))`
= `int (x"d"x)/(x^2sqrt(x^4 - 1))`
Put x2 = sec θ
∴ 2x dx = sec θ tan θ dθ
x dx = `1/2 sec theta tan theta "d"theta`
∴ I = `1/2 int (sec theta tan theta)/(sec theta sqrt(sec^2theta - 1)) "d"theta`
= `1/2 int (sectheta tan theta)/(sectheta * tan theta) "d"theta`
= `1/2 int 1 "d"theta`
= `1/2 theta + "C"`
So I = `1/2 sec^-1x^2 + "C"`
Hence, I = `1/2 sec^-1 x^2 + "C"`.
APPEARS IN
संबंधित प्रश्न
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integral:
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
