Question

$\int \frac{1}{e^{2\theta}+e^{-2\theta}} d\theta=$

• A. $\frac{1}{2} tan^{-1}\left(e^{2\theta}\right)+C$
• B. $\frac{1}{2} tan^{-1}\left(e^{-2\theta}\right)+C$
• C. $\frac{3}{2} tan^{-1}\left(e^{2\theta}\right)+C$
• D. $\frac{1}{2} tan^{-1}\left(-e^{2\theta}\right)+C$

Answer 1

Answer: (A)

$\frac{1}{2} tan^{-1}\left(e^{2\theta}\right)+C$

Answer 2

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 $e^{2\theta} = t$
$\Rightarrow e^{2\theta} \times 2\,d\theta = dt$
$\therefore I = \frac{1}{2}\int \frac{dt}{t^2+1} = \frac{1}{2} tan^{-1}(t)+C$
$=\frac{1}{2} tan^{-1} (e^{2\theta})+C$