Question:
Evaluate : $\int\frac{1}{sin \,x +\sqrt{3} \,cos\, x} dx$
Evaluate : $\int\frac{1}{sin \,x +\sqrt{3} \,cos\, x} dx$
Updated On: Jul 6, 2022
- $log \left|tan \left(\frac{x}{2}\right)\right| + C$
- $-\frac{1}{2}log \left|tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right| + C$
- $\frac{1}{2}log \left|tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right| + C$
- None of these
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The Correct Option is C
Solution and Explanation
Let $I = \int\frac{1}{sin\, x +\sqrt{3} cos \,x} dx$
$ = \frac{1}{2}\int\frac{dx}{\frac{1}{2}sin\, x + \frac{\sqrt{3}}{2} cos \,x}$
$\Rightarrow I= \frac{1}{2 }\int\frac{1}{sin \left(x+\frac{\pi}{3}\right)} dx = \frac{1}{2} \int cosec \left(x+\frac{\pi}{3}\right) dx $
$\Rightarrow I = \frac{1}{2}log \left|tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right| + C$
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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.