Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) d x\]
\[Here, f\left( x \right) = log\left( \frac{2 - \sin x}{2 + \sin x} \right)\]
\[f\left( - x \right) = log\left( \frac{2 - \sin\left( - x \right)}{2 + \sin\left( - x \right)} \right) = log\left( \frac{2 + \sin x}{2 - \sin x} \right) = - log\left( \frac{2 - \sin x}{2 + \sin x} \right) = - f\left( x \right)\]
\[\text{Hence} f\left( x \right) \text{is an odd function}\]
\[ \therefore I = 0\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
If f(2a − x) = −f(x), prove that
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Prove that:
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
