Question:
$\int _{\pi/3}^{\pi/3} \frac {x \sin x}{\cos^2x} dx $ is equal to
$\int _{\pi/3}^{\pi/3} \frac {x \sin x}{\cos^2x} dx $ is equal to
Updated On: Jul 6, 2022
- $\frac{\pi}{3}-\log\, \tan \frac{3\pi}{2}$
- $2[\frac {2\pi} {3}-\log\, \tan \frac {5\pi} {12}]$
- $3[\frac{\pi}{2}\log\, \sin \frac{\pi}{12}]$
- none of these
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The Correct Option is B
Solution and Explanation
$\int\limits_{-\pi /3}^{\pi /3} \frac{x\, sin\,x}{cos^{2}\,x} dx=2 $
$\int\limits_{0}^{\pi /3} x \left(sec\,x \, tan\,x\right)dx$
$=2\left[\left|x\cdot sec x\right|_{0}^{\pi /3}-\int_{0}^{\pi /3} sec\,x\,dx\right]$
$=2\left[\frac{\pi}{3}\cdot\left(2\right)-\left|log\left(sec\,x+tan\,x\right)\right|_{0}^{\pi /3}\right]$
$=2\left[\frac{2\pi}{3}-log\left(2+\sqrt{3}\right)+0\right]$
$=2\left[\frac{2\pi}{3}-log\, tan \frac{5\pi}{12}\right]$
$\left[\because tan \frac{5\pi}{12}=tan\,75^{\circ}=2+\sqrt{3}\right]$
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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.