Question:
$\int\limits_{-\pi/4}^{\pi/4}\frac{e^x\,\sec^2x\,dx}{e^{2x}-1}$ is equal to
$\int\limits_{-\pi/4}^{\pi/4}\frac{e^x\,\sec^2x\,dx}{e^{2x}-1}$ is equal to
Updated On: Jul 6, 2022
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- e
- none of these.
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The Correct Option is A
Solution and Explanation
Let $f\left(x\right) =\frac{e^{x} sec^{2}\,x}{e^{2x}-1}$
$\therefore f \left(-x\right)=\frac{e^{-x} sec^{2}\left(-x\right)}{e^{-2x}-1}=\frac{\frac{1}{e^{x}}}{\frac{1}{e^{2x}}-1}sec^{2}\, x$
$=\frac{e^{2} x}{e^{x}\left(1-e^{2x}\right)} sec^{2}\,x=-\frac{e^{x}}{e^{2x}-1}sec^{2}\, x=f \left(x\right)$
$\therefore$ f i s an odd function
$\therefore \int\limits_{-\pi 4}^{\pi 4}\frac{e^{x}sec^{2}\,x}{e^{2x}-1} dx=0$
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.