Advertisements
Advertisements
Question
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
Advertisements
Solution
\[\int_\frac{- 1}{2}^\frac{1}{2} \cos x \log\left( \frac{1 + x}{1 - x} \right) d x\]
\[\text{Let }f(x) = \cos x \log\left( \frac{1 + x}{1 - x} \right)\]
\[\text{Consider }f(- x) = \cos\left( - x \right) \log\left( \frac{1 - x}{1 + x} \right)\]
\[ = \cos x\left\{ - \log\left( \frac{1 + x}{1 - x} \right) \right\} = - \cos x \log\left( \frac{1 + x}{1 - x} \right) = - f\left( x \right)\]
Thus f(x) is an odd function
Therefore,
\[ \int_\frac{- 1}{2}^\frac{1}{2} \cos x \log\left( \frac{1 + x}{1 - x} \right) d x = 0\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
Prove that:
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
