Advertisements
Advertisements
प्रश्न
Advertisements
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
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate the following integral:
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Solve each of the following integral:
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_2^3 e^{- x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
