Question:
$\int_{\log\frac{1} {2}} ^{\log2} \, \sin (\frac {e^x-1} {e^x+1})dx$ is equal to
$\int_{\log\frac{1} {2}} ^{\log2} \, \sin (\frac {e^x-1} {e^x+1})dx$ is equal to
Updated On: Jul 6, 2022
- $\cos \frac {1} {3}$
- 0
- $2 \cos 2$
- none of these
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The Correct Option is B
Solution and Explanation
Let $f \left(x\right)=sin \left(\frac{e^{x}-1}{e^{x}+1}\right)$
$\therefore f\left(-x\right)=sin\left(\frac{e^{-x}-1}{e^{-x}+1}\right)=sin \left(\frac{\frac{1}{e^{x}}-1}{\frac{1}{e^{x}}+1}\right)$
$=sin \left(\frac{1-e^{x}}{1+e^{x}}\right)=-sin \left(\frac{e^{x}-1}{e^{x}+1}\right)=-f \left(x\right)$
$\therefore f \left(x\right)$ is an odd function.
$\therefore$ given integral $=\int_{-log\,2}^{log\,2} sin \left(\frac{e^{x}-1}{e^{x}+1}\right)dx=0$
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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.