Advertisements
Advertisements
प्रश्न
उत्तर
\[Let\, I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right) d \theta\]
\[Here\, f\left( \theta \right) = \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)\]
\[Consider\, f\left( - \theta \right) = \log\left[ \frac{a - \sin\left( - \theta \right)}{a + \sin\left( - \theta \right)} \right] = - \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right) = - f\left( \theta \right)\]
\[i . e . , f\left( \theta \right) \text{is odd function} . \]
\[\text{Therefore}, I = 0\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate each of the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
`int_0^(2a)f(x)dx`
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`