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 the following integral:
Write the coefficient a, b, c of which the value of the integral
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
Γ(1) is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
Find: `int logx/(1 + log x)^2 dx`
Evaluate \[\int_{0}^{1}x\,dx\].
What is an antiderivative of \[2x+1\]?
