Evaluate the definite integral:
$$\int_{\frac{\pi}{8}}^\frac{\pi}{4}(\csc(2\theta)-\cot(2\theta)\ d\theta$$
Finding the derivative gives me this, which is confirmed by the steps in Wolfram Alpha. This is the answer as the last step that I also got.
$$-2\csc(2\theta)\cot(2\theta)+2\csc^2(2\theta)$$
But then at the top, Wolfram Alpha says the answer is this:
$$\sec^2\theta$$
How did they get that?
Edit: I just realized that I'm solving the question wrong; I'm supposed to find the antiderivative and not the derivate. Either way, I wanted to know how the identity was found.
$\endgroup$ 42 Answers
$\begingroup$Steps are: \begin{equation} -2\csc(2\theta)\cot(2\theta) + 2\csc^{2}(2\theta) \end{equation} \begin{equation} -2* \frac{1}{2\sin(\theta)cos(\theta)} * \frac{cos(2\theta)}{sin(2\theta)} + 2\csc^{2}(2\theta) \end{equation} Simplify. \begin{equation} -2 * \frac{1-2\sin^{2}(\theta)}{\sin^{2}(2\theta)} + 2\csc^{2}(2\theta) \end{equation} Then \begin{equation} \frac{-2}{\sin^{2}(2\theta)} + \frac{1}{\cos^{2}(\theta)} + 2\csc^{2}(2\theta) \end{equation} \begin{equation} \frac{-2}{\sin^{2}(2\theta)} + \frac{1}{\cos^{2}(\theta)} + \frac{2}{\sin^{2}(2\theta)} \end{equation} Clearly, the -2 and +2 terms cancel, and you're left with $\frac{1}{\cos^{2}(\theta)}$ which is the same as $\sec^{2}(\theta)$
$\endgroup$ 4 $\begingroup$Notice, $$\int_{\pi/8}^{\pi/4}(\csc(2\theta)-\cot(2\theta))\ d\theta$$ $$=\frac{1}{2}\int_{\pi/8}^{\pi/4}(\csc(2\theta)-\cot(2\theta))\ d(2\theta)$$ $$=\frac{1}{2}\left(\ln|\csc (2\theta)-\cot (2\theta)|-\ln|\sin (2\theta)|\right)_{\pi/8}^{\pi/4}$$ $$=\frac{1}{2}\left(\ln\left|\frac{\csc (2\theta)-\cot (2\theta)}{\sin (2\theta)}\right|\right)_{\pi/8}^{\pi/4}$$ $$=\frac{1}{2}\left(\ln\left|\frac{\csc \frac{\pi}{2}-\cot \frac{\pi}{2}}{\sin \frac{\pi}{2}}\right|-\ln\left|\frac{\csc \frac{\pi}{4}-\cot \frac{\pi}{4}}{\sin \frac{\pi}{4}}\right|\right)$$ $$=\frac{1}{2}\left(\ln\left|\frac{1-0}{1}\right|-\ln\left|\frac{\sqrt {2}-1}{\frac{1}{\sqrt 2}}\right|\right)$$ $$=\frac{1}{2}\left(0-\ln\left|2-\sqrt 2\right|\right)$$ $$=\color{}{-\frac{1}{2}\ln\left(2-\sqrt 2\right)}=\color{red}{\frac{1}{2}\ln\left(\frac{2+\sqrt 2}{2}\right)}$$$$
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