Question:
$ \int\left(sec\,x\right)log \left(sec\,x-tan\,x\right)dx= $
$ \int\left(sec\,x\right)log \left(sec\,x-tan\,x\right)dx= $
Updated On: Jun 23, 2024
- $ \frac{1}{2}\left\{log\left(sec\,x-tan\,x\right)\right\}^{2}+C $
- $ -\frac{1}{2}\left\{log\left(sec\,x-tan\,x\right)\right\}^{2}+C $
- $ -\frac{3}{2}\left\{log\left(sec\,x-tan\,x\right)\right\}^{2}+C $
- $ -\frac{1}{2}\left\{log\left(sec\,x-tan\,x\right)\right\}^{-2}+C $
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The Correct Option is B
Solution and Explanation
Let $I = \int sec\,x \cdot log(sec\,x - tan\, x)dx$
Put $log(sec\,x - tan\,x) = t$
$\Rightarrow \frac{1}{sec\,x - tan\,x}\times (sec\,x \,tan\,x - sec^2\,x)dx = dt$
$\Rightarrow - sec\,x\,dx = dt$
$\therefore I = -\int t \,dt = \frac{-t^2}{2} + C$
$ = \frac{-1}{2}[log(sec\,x - tan\,x)]^2 + C$
Put $log(sec\,x - tan\,x) = t$
$\Rightarrow \frac{1}{sec\,x - tan\,x}\times (sec\,x \,tan\,x - sec^2\,x)dx = dt$
$\Rightarrow - sec\,x\,dx = dt$
$\therefore I = -\int t \,dt = \frac{-t^2}{2} + C$
$ = \frac{-1}{2}[log(sec\,x - tan\,x)]^2 + C$
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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.
