Question:

$ \int\limits_{0} ^{\pi/2}\frac{\cos\,2x\,dx}{(\sin\,x+\cos\,x)^2}$ is equal to

Updated On: Jul 6, 2022
  • 0
  • $\frac{\pi}{4}$
  • $\frac{\pi}{2}$
  • none of these.
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The Correct Option is A

Solution and Explanation

$\int\limits_{0}^{\pi /2} \frac{cos\,2x}{\left(sin\,x+cos\,x\right)^{2}} dx$ $=\int\limits_{0}^{\pi /2} \frac{cos^{2}\,x-sin^{2}\,x}{\left(sin\,x+cos\,x\right)^{2}}dx$ $=\int\limits_{0}^{\pi/ 2}\frac{cos\,x-sin\,x}{cos\,x+sin\,x}dx=\left|log \left(cos\,x+sin\,x\right)\right|_{0}^{\pi/ 2}$ $=log\,1-log\,1=0$
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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:

  1. The problem of getting a function if its derivative is given.
  2. The problem of getting the area bounded by the graph of a function under given situations.