Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) d x\]
\[Here, f\left( x \right) = log\left( \frac{2 - \sin x}{2 + \sin x} \right)\]
\[f\left( - x \right) = log\left( \frac{2 - \sin\left( - x \right)}{2 + \sin\left( - x \right)} \right) = log\left( \frac{2 + \sin x}{2 - \sin x} \right) = - log\left( \frac{2 - \sin x}{2 + \sin x} \right) = - f\left( x \right)\]
\[\text{Hence} f\left( x \right) \text{is an odd function}\]
\[ \therefore I = 0\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
Evaluate each of the following integral:
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
Find : `∫_a^b logx/x` dx
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_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
