Question:
$ \int \frac{1}{e^{2\theta}+e^{-2\theta}} d\theta= $
$ \int \frac{1}{e^{2\theta}+e^{-2\theta}} d\theta= $
Updated On: Jun 23, 2024
- $ \frac{1}{2} tan^{-1}\left(e^{2\theta}\right)+C $
- $ \frac{1}{2} tan^{-1}\left(e^{-2\theta}\right)+C $
- $ \frac{3}{2} tan^{-1}\left(e^{2\theta}\right)+C $
- $ \frac{1}{2} tan^{-1}\left(-e^{2\theta}\right)+C $
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The Correct Option is A
Solution and Explanation
Let $I = \int \frac{1}{e^{2\theta} + e^{-2\theta}} d\theta$
$= \int \frac{e^{2\theta}}{(e^{2\theta})^2 +1 } d\theta$
Put $e^{2\theta} = t $
$\Rightarrow e^{2\theta} \times 2\,d\theta = dt$
$\therefore I = \frac{1}{2}\int \frac{dt}{t^2+1} = \frac{1}{2} tan^{-1}(t)+C$
$=\frac{1}{2} tan^{-1} (e^{2\theta})+C$
$= \int \frac{e^{2\theta}}{(e^{2\theta})^2 +1 } d\theta$
Put $e^{2\theta} = t $
$\Rightarrow e^{2\theta} \times 2\,d\theta = dt$
$\therefore I = \frac{1}{2}\int \frac{dt}{t^2+1} = \frac{1}{2} tan^{-1}(t)+C$
$=\frac{1}{2} tan^{-1} (e^{2\theta})+C$
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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.
