Advertisements
Advertisements
Question
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
Advertisements
Solution
\[\int_0^1 \sqrt{\frac{1 - x}{1 + x}} d x\]
\[ = \int_0^1 \sqrt{\frac{1 - x}{1 + x} \times \frac{1 - x}{1 - x}} d x\]
\[ = \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} d x\]
\[ = \int_0^1 \frac{1}{\sqrt{1 - x^2}}dx - \int_0^1 \frac{x}{\sqrt{1 - x^2}}dx\]
\[ = \left[ \sin^{- 1} x \right]_0^1 + \left[ \sqrt{1 - x^2} \right]_0^1 \]
\[ = \frac{\pi}{2} - 1\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Prove that:
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "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^oo "e"^(-2x) "d"x` is
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
