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