Question:
If $\int \frac{f\left(x\right)}{log\,cos\,x}dx=-log\left(log\,cos\,x\right)+C$, then $f\left(x\right)$ is equal to
If $\int \frac{f\left(x\right)}{log\,cos\,x}dx=-log\left(log\,cos\,x\right)+C$, then $f\left(x\right)$ is equal to
Updated On: May 18, 2024
- $tan\,x$
- $-sin\,x$
- $-cos\,x$
- $-tan\,x$
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The Correct Option is A
Solution and Explanation
$\int \frac{f(x)}{\log\, \cos\, x} dx=-\log (\log \cos\, x)+C$
Differentiating on both sides w.r.t. $x$,
$\frac{d}{dx}\left\{\int \frac{f(x)}{\log \,\cos\, x} d x\right\} $
$=-\frac{d}{d x}\{\log (\log \cos x)\}+\frac{d}{d x}(C) $
$ \Rightarrow \, \frac{f(x)}{\log \cos x}=\frac{-1}{\log \cos x} \cdot \frac{1}{\cos x} \cdot(-\sin x)+0 $
$\Rightarrow\, \frac{f(x)}{\log \cos x}=\frac{\tan x}{\log \cos x} $
$\Rightarrow \, f (x) =\tan \,x $
Differentiating on both sides w.r.t. $x$,
$\frac{d}{dx}\left\{\int \frac{f(x)}{\log \,\cos\, x} d x\right\} $
$=-\frac{d}{d x}\{\log (\log \cos x)\}+\frac{d}{d x}(C) $
$ \Rightarrow \, \frac{f(x)}{\log \cos x}=\frac{-1}{\log \cos x} \cdot \frac{1}{\cos x} \cdot(-\sin x)+0 $
$\Rightarrow\, \frac{f(x)}{\log \cos x}=\frac{\tan x}{\log \cos x} $
$\Rightarrow \, f (x) =\tan \,x $
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
