Question:
$\int \frac{sec\,x\,dx}{\sqrt{cos\,2\,x}}$ is equal to
$\int \frac{sec\,x\,dx}{\sqrt{cos\,2\,x}}$ is equal to
Updated On: Jun 6, 2022
- $2\, \sin^{-1}\, (\tan\,x) + C$
- $\tan^{-1}\,\left(\frac{\tan\,x}{2}\right)+C$
- $\sin^{-1}\, (\tan\,x) + C$
- $\frac{1}{2}\sin^{-1}\,\left(\tan\,x\right)+C$
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The Correct Option is C
Solution and Explanation
Let $ I =\int \frac{\sec x}{\sqrt{\cos 2 x}} d x$
$=\int \frac{\sec x}{\sqrt{\frac{1-\tan ^{2} x}{1+\tan ^{2} x}}} d x$
$=\int \frac{\sec ^{2} x}{\sqrt{1-\tan ^{2} x}} d x$
Put $ \tan x=t $
$ \Rightarrow \sec ^{2} x d x =d t$
$\therefore I =\int \frac{d t}{\sqrt{1-t^{2}}} $
$=\sin ^{-1} t+C $
$=\sin ^{-1}(\tan x)+C $
$=\int \frac{\sec x}{\sqrt{\frac{1-\tan ^{2} x}{1+\tan ^{2} x}}} d x$
$=\int \frac{\sec ^{2} x}{\sqrt{1-\tan ^{2} x}} d x$
Put $ \tan x=t $
$ \Rightarrow \sec ^{2} x d x =d t$
$\therefore I =\int \frac{d t}{\sqrt{1-t^{2}}} $
$=\sin ^{-1} t+C $
$=\sin ^{-1}(\tan x)+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.
