$ \int\left[sin \left(log\,x\right)+cos\left(log\,x\right)\right]dx $ is equal to
- $ x\, cos (log\,x)+c $
- $ cos (log\,x)+c $
- $ x\, sin (log\,x)+c $
- $ sin (log\,x)+c $
The Correct Option is C
Approach Solution - 1
Approach Solution -2
Given:
\(\int [\sin(\log x) + \cos(\log x)] \, dx\)
To integrate \(\sin(\log x) + \cos(\log x)\), we can recognize that its derivative to x is \(\frac{d}{dx}[x \sin(\log x)]\):
\(\frac{d}{dx}[x \sin(\log x)] = \sin(\log x) + x \cdot \frac{d}{dx}[\sin(\log x)]\)
Now, differentiate \(\sin(\log x)\):
\(\frac{d}{dx}[\sin(\log x)] = \cos(\log x) \cdot \frac{1}{x}\)
So,
\(\frac{d}{dx}[x \sin(\log x)] = \sin(\log x) + x \cos(\log x) \cdot \frac{1}{x} = \sin(\log x) + \cos(\log x)\)
Therefore,
\(\int [\sin(\log x) + \cos(\log x)] \, dx = x \sin(\log x) + C\)
So, the correct option is (C): \(x \sin(\log x) + C\)
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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.