Question:
Evaluate : $ \int\limits_{0}^{\pi/2}\frac{ cos \,x}{\left(cos \frac{x}{2} +sin \frac{x}{2}\right)^{3}} dx$
Evaluate : $ \int\limits_{0}^{\pi/2}\frac{ cos \,x}{\left(cos \frac{x}{2} +sin \frac{x}{2}\right)^{3}} dx$
Updated On: Jul 6, 2022
- $2 -\sqrt{2}$
- $2 +\sqrt{2}$
- $3 +\sqrt{3}$
- $3 -\sqrt{3}$
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
We have,
$I = \int\limits_{0}^{\pi/2}\frac{ cos \,x}{\left(cos \frac{x}{2} +sin \frac{x}{2}\right)^{3}} dx $
$= \int\limits_{0}^{\pi /2} \frac{cos^{2 } \frac{x}{2} - sin^{2} \frac{x}{2} }{\left(cos \frac{x}{2} + sin \frac{x}{2}\right)^{3}} dx = \int\limits_{0}^{\pi /2} \frac{cos^{ } \frac{x}{2} - sin \frac{x}{2} }{\left(cos \frac{x}{2} + sin \frac{x}{2}\right)^{2}} dx $
Let $cos \frac{x}{2} +sin \frac{x}{2} = t$
$ \Rightarrow \frac{1}{2}\left(-sin \frac{x}{2} +cos \frac{x}{2}\right) dx = dt $
Also, $x = 0 \Rightarrow t = 1\,$ and $\, x = \frac{\pi}{2} \Rightarrow t = \sqrt{2}$
$\therefore I = \int\limits_{0}^{\sqrt{2}} \frac{2dt}{t^{2}} 2\left[-\frac{1}{t}\right]_{1}^{\sqrt{2}} $
$= 2\left[-\frac{1}{\sqrt{2}} +1\right] = \left(2-\sqrt{2}\right)$
Was this answer helpful?
0
0
Top Questions on integral
If \[ \int e^x (x^3 + x^2 - x + 4) \, dx = e^x f(x) + C, \] then \( f(1) \) is:
The value of : \( \int \frac{x + 1}{x(1 + xe^x)} dx \).
- The order and degree of the differential equation \( \sqrt{\frac{dy}{dx}} - 4 \frac{dy}{dx} - 7x = 0 \) are respectively
- The area enclosed by the curves \( y = x^3 \) and \( y = \sqrt{x} \) is:
- The value of \( \int e^{\tan \theta} (\sec \theta - \sin \theta) \, d\theta \) is:
View More Questions
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.